Chapter # 46
The Nucleus
SOLVED EXAMPLES
1. Sol.
Calculate the radius of 70Ge. We have, R = R0 A1/3 = (1.1 fm) (70)1/3 = (1.1 fm) (4.12) = 4.53 fm.
2.
Calculate the binding energy of an alpha particle from the following data: mass of 11H atom mass of neutron
Sol.
3. Sol. 4.
Sol.
= 1.007826 u = 1.008665 u
mass of 24 He atom = 4.00260 u Take 1 u = 931 MeV/c2. The alpha particle contains 2 protons and 2 nutrons. The binding energy is B = (2 × 1.007825 u + 2 × 1.008665 u – 4.00260 u)c2 = (0.03038 u)c2 = 0.03038 × 931 MeV = 28.3 MeV. The atomic mass 11H is 1.00783 u. Calculate the mass excess of hydrogen. The mass excess of hydrogen is 931 (m – A)MeV = 931(1.00783 – 1)MeV = 7.29 MeV. The decay constant for the radioactive nuclide 64Cu is 1.516 × 10–5 s–1. Find the activity of a sample containing 1 g of 64Cu.Atomic weight of copper = 63.5 g/mole. Neglect the mass difference between the given radioisotope and normal copper. 63.5 g of copper has 6 × 1023 atoms. Thus, the number of atoms in 1 g of Cu is N= The activity
6 10 23 1g = 9.45 × 1015) 63.5 g
= N = (1.516 × 10–5 s–1) × (9.45 × 1015) = 1.43 × 1011 disintegrations/s =
5. Sol.
1.43 1011 3.7 1010
Ci = 3.86 Ci.
The half-life of a radioactive nuclide is 20 hours. What fraction of original activity will remain after 40 hours? We have
1 40 hours t1/ 2 = 20 hours = 2. Thus,
A=
A0 2 t / t1 / 2
A0 22
A0 4
A 1 A0 = 4 . So one fourth of the original activity will remain after 40 hours. or,
6. Sol.
The binding energy per nucleon is 8.5 MeV for A 120 and is 7.6 MeV for A = 240. Suppose a nucleus with A = 240 breaks into two nuclei of nearly equal mass numbers. Calculate the energy released in the process. Suppose the heavy nucleus had Z protons and N neutrons. The rest mass energy of this nucleus would be E = Mc2 = (Zmp + Nmn)c2 – B1 = (Zmp + Nmn)c2 – 7.6 × 240 MeV. If there are Z1 protons and N1 neutrons in the first fragment, its rest mass energy will be E1 = M1c2 = (Z1mp + N1mn)c2 – B2 = (Z1mp + N1mn)c2 – (8.5 MeV) (Z1 + N1). Similarly, if there are Z2 protons and N2 neutrons in the first fragment, its rest mass energy will be E2 = (Z2mp + N2mn)c2 – (8.5 MeV) (Z2 + N2). The energy released due to the breaking is E – (E1 + E2) = [(Z – Z1 – Z2]mp c2 + (N – N1 – N2)mnc2] + [(Z1 + Z2 + N1 + N2) × 8.5 – 240 × 7.6] MeV = 240 × (8.5 – 7.6) MeV = 216 MeV. We have used the fact that Z1 + Z2 = Z, N1 + N2 = N and Z1 + Z2 + N1 + N2 = Z + N = 240. Thus, 216 MeV of manishkumarphysics.in
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Chapter # 46 The Nucleus energy will be released when this nucleus breaks. 7.
Sol.
Consider two deuterons moving towards each other with equal speeds in a deutron gas. What should be their kinetic energies (when they are widely separated) so that the closest separation between them becomes 2fm? Assume that the nuclear force is not effective for separations greater than 2 fm. At what temperature will the deuterons have this kinetic energy on an average? As the deuterons move, the Coulomb repulsion will slow them down. The loss in kinetic energy will be equal to the gain in Coulomb potential energy. At the closest separation, the kinetic energy is zero and the
e2 potential energy is . If the initial kinetic energy of each deuteron is K and the closest separation is 4 0r 2fm, we shall have 2K =
=
e2 4 0 (2 fm) (1.6 10 19 C) 2 (9 10 9 N m 2 / C 2 ) 2 10 15 m
or, K = 5.7 × 10–14 J. If the temperature of the gas is T, the average kinetic energy of random motion of each nucleus will be 1.5 kT. The temperature needed for the deuterons to have the average kinetic energy of 5.7 × 10–14 J will be given by 1.5 kT = 5.7 × 10–14 J where k = Botzmann constant or,
T=
5.7 10 14 J
1.5 1.38 10 23 J / K = 2.8 × 109 K.
WORKD 1. Sol.
EXAMPLES
OUT
Calculate the electric potential energy due to the electric repulsion between two nuclei of 12C when they ‘touch’ each other at the surface The radius of a 12C nucleus is R = R0 A1/3 = (1.1 fm) (12)1/3 = 2.52 fm. The separation between the cntres of the nuclei is 2R = 5.04 fm. The potential energy of the pair is
q1q2 U = 4 r 0 = (9 × 109 N–m2/C2)
(6 1.6 10 19 C)2
5.04 10 15 m = 1.64 × 10–12 J = 10.2 MeV.
2.
Find the binding energy of
56 26Fe .
Atomic mass of
56
Fe is 55.9349 u and that of 1H is 1.00783 u. Mass of
neutron = 1.00867 u. Sol.
The number of protons in
56 26Fe
= 26 and the number of neutrons = 56 – 26 = 30. The binding energy of
56 26Fe
is = [26 × 1.00783 u + 30 × 1.00867 u – 55.9349 u] c2 = (0.52878 u)c2 = (0.52878 u) (931 MeV/u) = 492 MeV. 3.
Find the kinetic energy of the -particle emitted in the decay
238
Pu 234 U . The atomic masses needed
are as follows : 238
234 Pu U 238.04955 u 234.04095 u Neglect any recoil of the residual nucleus.
Sol.
4
He 4.002603 u
Using energy conservation, m(238Pu)c2 = m (234U)c2 + m(4He)c2 + K manishkumarphysics.in
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Chapter # 46 The Nucleus or, K = [m(238Pu) – m(234U) – m(4He)]c2 = [238.04955 u – 234.04095 u – 4.002603 u] (931 Me V/u) = 5.58 MeV. 4.
Sol.
Calculate the Q-value in the following decays : (a) 19O 19F + e + (b) 25Al 25Mg + e+ + . The atomic masses needed are as follows: 19 19 25 O F Al 19.003576 u 18.998403 u 24.990432 u (a) The Q-value of – -decay is Q = [m(19O) – m(19F)]c2 = [19.003576 u – 18.998403 u ] (931 MeV/u) = 4.816 MeV (b) The Q-value of + -decay is Q = [m(25Al) – m(25Mg) – 2me]c2
25
Mg 24.985839 u
MeV 2 = 24.99032 u 24.985839 u 2 0.511 2 c c = (0.004593 u) (931 MeV/u) – 1.022 MeV = 4.276 MeV – 1.022 MeV = 3.254 MeV.
5.
Sol.
6.
Sol.
Find the maximum energy that a beta particle can have in the following decay 176 Lu 176 Hf + e + 176 Atomic mass of Lu is 175.942694 u and that of 176Hf is 175.941420 u. The kinetic energy available for the beta particle and the antineutrino is Q = [m(176Lu) – m (176Hf)]c2 = (175.942694 u – 175.941420 u) (931 MeV/u) = 1.182 MeV. This energy is shared by the beta particle and the antineutrino. The maximum kinetic energy of a beta particle in this decay is, therefore, 1.182 MeV when the antineutrino practically does not get any share. Consider the beta decay 198 Au 198 Hg* + – + 198 where Hg* represents a mercury nucleus in an excited state at energy 1.088 MeV above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass 198Au is 197.968233 u and that of 198Hg is 197.966760 u. If the product nucleus 198Hg is formed in its ground state, the kinetic energy available to the electron and the antineutrino is Q = [m(198Au) – m(198Hg)]c2 . 198 As Hg* has energy 1.088 MeV more than 198Hg in ground state, the kinetic energy actually available is Q = [m(198Au) – m(198Hg)]c2 – 1.088 MeV MeV – 1.088 MeV = (197.968233 u – 197.966760 u) 931 u = 1.3686 MeV – 1.088 MeV = 0.2806 MeV. This is also the maximum possible kinetic energy of the electron emitted.
7.
Sol.
The half-life of 198Au is 2.7 days. Calculate (a) the decay constant, (b) the average-life and (c) the activity of 1.00 mg of 198Au. Take atomic weight of 198Au to be 198 g/mol. 198 Au dk v)Z&vk;qdky 2.7 fnu gSA crkb;s (a) {k; fu;rkad (b) vkSlr vk;q (c) 1.00 mg 198Au dh lfØ;rkA 198Au dk ijek.kq Hkkj 198 g/mol yhft,A (a) The half-life and the decay constant are related as t1/2 = or,
=
ln 2 0.693 =
0.693 0.693 t1/ 2 = 2.7 days
0.693 = 2.7 24 3600 s = 2.9 × 10–6 s–1. manishkumarphysics.in
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Chapter # 46 (b) (c)
The Nucleus 1 = 3.9 days. The activity is A = N. Now, 198 g of 198Au has 6 × 1023 atoms. The number of atoms in 1.00 mg of 198 Au is
The avergae-life is tav =
1.0 mg N = 6 × 1023 × 198 g = 3.03 × 10 18. A = N = (2.9 × 10–6 s–1) (3.03 × 10 18) = 8.8 × 1012 disintegrations/s
Thus,
= 8. Sol.
8.8 1012 3.7 1010
Ci = 240 Ci.
A radiactive sample has 6.0 × 1018 active nuclei at a certain instant. How many of these nuclei will still be in the same active state after two half-lives? In one half-life the number of active nuclei reduces to half the original number. Thus, in two half lives the 1 1 number is reduced to of the original number. The number of remaining active nuclei is, therefore, 2 2
1 1 6.0 × 1018 × 2 2 18 = 1.5 × 10 .
9.
The activity of a radioactive sample falls from 600 s –1 to 500 s –1 in 40 minutes. Calculate its half-life.
fdlh jsfM;ks/kehZ inkFkZ dh lfØ;rk
40 fefuV
esa
, 600 s –1 ls 500 s –1 fxj
tkrh gSA bldh v)Z&vk;q dh x.kuk dhft,& HCV_Ch-46_WOE_9
Sol.
We have, or,
A = A0 e–t 500 s–1 = (600 s–1) e–t
or,
e–t =
or,
5 6 t = ln(6/5)
or,
=
The half-life is t1/2 =
ln(6 / 5) ln( 6 / 5) = 40 min t =
ln 2
ln 2 × 40 min ln(6 / 5)
= 152 min. 10.
Sol.
11.
Sol.
The number of 238U atoms in an ancient rock equals the number of 206Pb atoms. The half-life of decay of 238U is 4.5 × 10 9 y. Estimate the age of the rock assuming that all the 206Pb atoms are formed from the decay of 238 U. Since the number of 206Pb atoms equals the number of 238U atoms, half of the original 238U atoms have decayed. It takes one half-life to decay half of the active nuclei. Thus, the sample is 4.5 × 109 y old. Equal masses of two samples of charcoal A and B are burnt separately and the resulting carbon dioxide are collected in two vessels. The radioactivity of 14C is measured for both the gas samples. The gas from the charcoal A gives 2100 counts per weak and the gas from the charcoal B gives 1400 counts per week. Find the age difference between the two samples. Half-life of 14C = 5730 y. The activity of sample A is 2100 counts per week. After a cetain time t, its activity will be reduced to 1400 counts per week. This is because a fraction of the active 14C nuclei will decay in time t. The sampel B must be a time t older than the sample A. We have, A = A0 e – t –1 or, 1400 s = 2100 s–1 e–t manishkumarphysics.in
Page # 4
Chapter # 46 or,
The Nucleus e–t = t=
12.
Sol.
13.
Sol.
2 3
ln(3 / 2)
=
ln(3 / 2) t 0.693 1/2
=
0.4055 × 5730 y = 3352 y.. 0.693
Suppose, the daughter nucleus in a nuclear decay is itself radioactive. Let p and d be the decay constants of the parent and the daughter nuclei. Also, let Np and Nd be the number of parent and daughter nuclei at time t. Find the condition for which the number of daughter nuclei becomes constant. The number of parent nuclei decaying in a short time interval t to t + dl is p Npdt. This is also the number of daughter nuclei decaying during the same time interval is dNddt. The number of the daughter nuclei will be constant if pNpdt = dNddt or, pNp = dNd. A radioactive sample decays with an avergae-life of 20 ms. A capacitor of capcitance 100 F is charged to some potential and then the plates are connected through a resistance R. What should be the value of R so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time? The activity of the sample at time t is given by A = A0 e–t where is the decay constant and A0 is the activity at time t = 0 when the capacitor at time t is given by Q = Q0 e–t/CR where Q0 is the charge at t = 0 and C = 100 F is the capacitance. Thus,
Q Q 0 e t / CR . A A 0 e t
14.
It is independent of t if
=
1 CR
or,
R=
t av 1 20 10 3 s = = = 200 . C C 100 10 6 F
A radioactive nucleus can decay by two different processes. The half-life for the first process is t1 and that for the second process is t2. Show that the effective half-life t of the nucleus is given by
1 1 1 t t1 t 2 . Sol.
The decay constant for the first process is 1 =
ln 2 ln 2 and for the second process it is = 2 t1 t1 . The
probability that an active nucleus decays by the first process in a time interval dt is 1dt. Similarly, the probability that it decays by the second process is 2dt. The probability that it either decays by the first process or by the scond process is 1dt + 2dt. If the effective decay constant is , this probability is also equal to dt. Thus. dt = 1dt + 2dt or, = 1 + 2 or, 15.
1 1 1 t t1 t 2 .
Calculate the energy released when three alpha particles combine to form a 12C nucleus. The atomic mass of 24He is 4.002603 u.
Sol.
4 12 The mass of a 12C atom is exactly 12 u. The energy released in the reaction 3 2 He 6 C is
manishkumarphysics.in
Page # 5
Chapter # 46
The Nucleus 12 [3 m( 24 He ) – m( 6 C )] c2
= [3 × 4.002603 u – 12 u] (931 MeV/u) = 7.27 MeV.
Question for short answer 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
If neutrons exert only attractive force, why don’t we have a nucleus containing neutrons alone ? Consider two pairs of neutrons. In each pair, the separation between the neutrons is the same. Can the force between the neutrons have different magnitudes for the two pairs ? A molecule of hydrogen contains two protons and two electrons. The nuclear force between these two protons is always neglected while discussing the behaviour of a hydrogen molecule. Why ? Is it easier to take out a nucleon from carbon or from iron ? From iron or from lead ? Suppose we have 12 protons and 12 neutrons. We can assemble them to from either a 24Mg nucleus or two 12 C nuclei. In which of the two cases more energy will be liberated ? What is the difference between cathode rays and beta rays ? When the two are travelling in space, can you make out which is the cathode ray and which is the beta ray ? [HCV_Chp. 46_Q.Short A_6] If the nucleons of a nucleus are separated from each other, the total mass is increased, Where does this mass come from ? In beta decay, an electron (or a positron) is emitted by a nucleus. Does the remaining atom get oppositely charged ? [HCV_Chp. 46_Q.Short A_8] 10 When a boron nucleus ( 5 B ) is bombarded by a neutron, an -particle is emitted. Which nucleus will be formed as a result ? Does a nucleus lose mass when it suffers gamma decay ? In a typical fission reaction, the nucleus is split into two middle-weight nuclei of unequal masses. Which of the two (heavier or lighter) has greater kinetic energy ? Greater linear momentum ? If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can’t helium nuclei combine on their own and minimise the energy ?
Objective - I 1.
The mass of a netural carbon atom in ground state is (A*) exact 12 u (B) less than 12 u (C) more than 12 u (D) depends on the from of carbon such as graphite or charcoal. mnklhu dkcZu ijek.kq dk ewy voLFkk esa nzO;eku gS (A*) Bhd 12 u (B) 12 u ls de (C) 12 u ls vf/kd (D) dkcZu dh voLfkk ij fuHkZj djsxk tSls xzsQkbV ;k pkjdksy
2.
The mass number of a nucleus is equal to (A) the number of neutrons in the nucleus (C*) the number of protons in the nucleus ukfHkd dh nzO;eku la[;k rqY; gksrh gS (A) ukfHkd esa U;wVªkWuksa dh la[;k ds cjkcj (C*) ukfHkd esa U;wfDyvkWuksa dh la[;k ds cjkcj
(B) the number of protons in the nucleus (D) none of them (B) ukfHkd esa izksVkWuksa dh (D) buesa ls dksbZ ugha
la[;k ds cjkcj
3.
As compared to 12C atom, 14C atom has (A) two extra protons and two extra electrons (B) two extra protons but no extra electron (C*) two extra neutorns and no extra electrons (D) two extra neutons and two extra electrons 12 14 C ijek.kq dh rqyuk esa] C ijek.kq esa gksrs gSa (A) nks vfrfjDr izkVs kWu rFkk nks vfrfjDr bysDVªkWu (B) nks vfrfjDr izkVs kWu fdUrq dksbZ vfrfjDr bysDVªkWu ugha (C*) nks vfrfjDr U;wVªkWu fdUrq dksbZ vfrfjDr bysDVªkWu ugha (D) nks vfrfjDr U;wVªkWu rFkk nks vfrfjDr bysDVªkWu
4.
The mass number of a nucleus is (A) always less than its atomic number (B) always more than its atomic number (C) equal to its atomic number (D*) sometimes more than and sometimes equal to its atomic number fdlh ukfHkd dh nzO;eku la[;k lnSo (A) blds ijek.kq Øekad ls de gksrh gSA (B) blds ijek.kq Øekad ls vf/kd gksrh gSA manishkumarphysics.in
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Chapter # 46 (C) blds ijek.kq (D*) blds ijek.kq 5.
The Nucleus
Øekad ds cjkcj Øekad ls dHkh de o dHkh vf/kd
The graph of n (R/R0) versus In A(R=radius of a nucleus and A = its mass number) is (A*) a straight line (B) a parabola (C) an ellipse (D) none of them n (R/R0) o n A ds lkis{k xzkQ (R=fdlh ukfHkd dh f=kT;k ,oa A = bldh nzO;eku la[;k½ gksrk (A*) ljy js[kk (B) ijoy; (C) nh?kZoÙ` k (D) buesa ls dksbZ ugha
gS -
6.
Let Fpp, Fpn and Fnn denote the magnitudes of the nuclear force by a proton on a proton, by a proton on a neutron and by a neutron on a neutron respectively. When the separation is 1 fm, ekuk fd Fpp, Fpn vkSj Fnn Øe'k% izksVkWu }kjk izkVs kWu ij] izkVs kWu }kjk U;wVkª Wu ij ukfHkdh; cy dk ifjek.k O;Dr djrs gSaA tc nwjh 1 fm ¼QehZ½ gks rks (A) Fpp> Fpn = Fnn (B*) Fpp= Fpn = Fnn (C) Fpp> Fpn > Fnn (D) Fpp< Fpn = Fnn
7.
Let Fpp, Fpn and Fnn denote the magnitudes of the net force by a proton on a proton by a proton on a neutron and by a neutron on a neutron respectively. Neglect gravitational force. When the separation is 1 fm, ekuk fd Fpp, Fpn vkSj Fnn Øe'k% izksVkWu }kjk izkVs kWu ij] izkVs kWu }kjk U;wVkª Wu ij ukfHkdh; cy dk ifjek.k O;Dr djrs gSaA xq:Roh; cy dks ux.; eku yht;sA tc nwjh 1 fm¼QehZ½ gks rks (A) Fpp> Fpn = Fnn (B) Fpp= Fpn = Fnn (C) Fpp> Fpn > Fnn (D*) Fpp< Fpn = Fnn
8.
Two protons are kept at a separation of 10 nm. Let Fn and Fe be the nuclear force and the electromagnetic force between them. nks izkVs kWuksa dh 10 nm nwjh ij j[kk x;k gSA ekuk fd buds chp ukfHkdh; cy ,oa fo|qr&pqEcdh; cy Fn o Fe gS] rks (A) Fe = Fn (B*) Fe >> Fn (C) Fe << Fn (D) Fe o Fn
9.
As the mass number A increases, the binding energy per nucleon in a nucleus (A) increases (B) decreases (C) remains the same (D*) varies in a way the depends on the actual value of A. tSls&tSls nzO;eku la[;k A c<+rh gS] ukfHkd esa izfr U;wfDyvkWu cU/ku ÅtkZ (A) c<+rh gSA (B) de gksrh gSA (C) ogh jgrh gSA (D*) bl izdkj ifjofrZr gksrh gS fd bldk eku A ij fuHkZj djrk gSA
10.
Which of the following is a wrong description of binding energy of a nucleus ? (A) It is the energy required to break a nucleus into its constituent nucleons. (B) It is the energy mad avilable when free nucleous combine to from a nucleus (C) It is the sum of the rest mass energies of its nucleous minus the rest mass energy of the nucleus (D*) It is the sum of the kinetic energy of all the nucleous in the nucleus ukfHkd dh ca/ku ÅtkZ ds fy;s fuEu esa ls dkSulh O;k[;k xyr gS (A) ukfHkd dks blds cukus okys U;wfDyvkWuksa esa rksM+us ds fy;s vko';d ÅtkZ gSA (B) ;g og ÅtkZ gS tks Lora=k U;wfDyvkWuksa dks la;ksftr djds ukfHkd cukus ij eqDr gksrh gSA (C) ;g blds U;wfDyvkWuksa dh fojke ÅtkZ dks ?kVkus ij izkIr gksrh gSA (D*) ;g ukfHkd esa leLr U;wfDyvkWuksa dh xfrt ÅtkZvksa dk ;ksx gSA
11.
In one average-life (A) half the active nuclei decay (C*) more than half the active nuclei decay ,d vkSlr&vk;q esa (A) vk/ks lfØ; ijek.kq fo?kfVr gks tkrs gSAa (C*) vk/ks ls vf/kd ijek.kq fo?kfVr gks tkrs gSAa
12.
(B) less than half the active nuclei decay (D) all the nuclei decay (B) vk/ks ls de ijek.kq (D) lkjs ukfHkd fo?kfVr
fo?kfVr gks tkrs gSAa gks tkrs gSAa
In a radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay ? (A) proton (B) neutorn (C) electron (D*) photon
fdlh jsfM;ks lfØ; fo[k.Mu esa u rks ijek.kq Øekad ifjofrZr gksrk gS] vkSj u gh nzO;eku la[;kA bl fo[k.Mu esa fuEu d.kksa esa ls dkSulk mRlftZr gksrk gS (A) izksVkWu (B) U;wVªkWu (C) bysDVªkWu (D*) QksVkWu 13.
During a negative beta decay, (A) an atomic electron is ejected (B) an electron which is already present within the nucleus is ejected (C*) a neutron in the nucleus decays emitted an electron manishkumarphysics.in
Page # 7
Chapter # 46
The Nucleus
(D) a proton in the nucleus decays emitting _.kkRed chVk {k; esa - , (A) ,d ijekf.od bysDVªkWu mRlftZr gksrk gSA (B) ukfHkd esa igys ls fo|eku ,d bysDVªkWu mRlftZr gksrk gSA (C*) ukfHkd esa U;wVªkWu ,d bysDVªkWu mRlftZr djds fo[kf.Mr gksrk gSA (D) ukfHkd esa izksVkWu ,d bysDVªkWu mRlftZr djds fo[kf.Mr gksrk gSA 14.
A freshly prepared radiocative source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is ,d rktk rS;kj fd;s x;s jsfM;kslfØ; lzkrs dh v)Z vk;q 2 ?kaVk gS] vkSj bldh rhozrk vuqer lqjf{kr Lrj dh 64 xquk gSA bl lzksr dk mi;ksx djus ds fy;s vko';d U;wure le; gS (A) 6 h (B*) 12 h (C) 24 h (D) 128 h
15.
The decay constant pf a radoactive sample is . The half-life and the average-life of the sample are respectively (A) 1/ and (In 2/) (B*) (In2/ and 1/ (C) 1(In 2) and 1/ (D) /(In 2) and 1/ ,d jsfM;kslfØ; izfrn'kZ dk {k; fu;rkad gSA bl izfrn'kZ dh v)Z&vk;q ,oa vkSlr vk;q dk eku Øe'k% gS (A) 1/ vkSj (In 2/) (B*) (In2/ vkSj 1/ (C) 1(In 2) vkSj 1/ (D) /(In 2) vkSj 1/
16.
An a-particle is bombarded on 14N. As a result, a 17O nucleus is formed and a particle is emitted. This particle is (A) neutron (B*) proton (C) electron (D) positron 14 17 N ij ,d d.k dh ceckjh dh tkrh gSA ifj.kkeLo:i ,d O ukfHkd curk gS rFkk ,d d.k mRlftZr gksrk gSA ;g d.k gS (A) U;wVªkWu (B*) izksVkWu (C) bysDVªkWu (D) ikWftVªkWu
17.
Ten grams of 57Co kept in an open container beta-decays with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearly (A*) 10 g (B) 5 g (C) 2.5 g (D) 1.25 g ,d [kqys ik=k esa j[ks gq, 10 xzke 57Co dk chVk {k; gksrk gS] ftldh v)Zvk;q 270 fnu gSA 540 fnu ds i'pkr~ ik=k ds vUnj inkFkZ dk Hkkj gksxk] yxHkx (A*) 10 xzke (B) 5 xzke (C) 2.5 xzke (D) 1.25 xzke
18.
Free 238U nuclei kept in a train emit alpha particles. When the train is stationery and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus beomes x in time t after the decay. If a decay takes place when the train is moving at a uniform speed , the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger will be (A) x + t (B) x - t (C*) x (D) depends on the direction of the train 238 Vªus esa j[kk gqvk ,d Lora=k U ukfHkd -d.k mRlftZr djrk gSA tc Vªus fLFkj gS o ;wjfs u;e ukfHkd fo[kf.Mr gksrk gS] ,d ;k=kh a-d.k rFkk izfrf{kIr ukfHkd ds chp dh nwjh ekirk gS] rks {k; ds 4 le; i'pkr~ x gks tkrh gSA ;fn fo[k.Mu rc gksrk gS] tc Vªus ,d leku pky ls py jgh gS] rks fo[k.Mu ds t le; ds i'pkr~ d.k rFkk izfrf{kIr ukfHkd ds chp ;k=kh }kjk ekih x;h nwjh gksxh (A) x + t (B) x - t (C*) x (D) Vªsu dh fn'kk ij fuHkZj djsxhA
19.
During a nuclear fission reaction, (A) a heavy nucleus breaks into two fragments by itself (B) a light nucleus bombarded by thermal neutrons break up (C*) a heavy nucleus bombarded by thermal neutrons breaks up (D) two light nuclei combine to give a heavier nucleus and possibly other products. ukfHkdh; fo[k.Mu vfHkfØ;k esa (A) ,d Hkkjh ukfHkd Lo;a gh nks Hkkxksa esa fo[kf.Mr gksrk gSA (B) gYds ukfHkd ij rkih; U;wVkª Wuksa dh ceckjh ls ;g VwV tkrk gSA (C*) Hkkjh ukfHkd ij rkih; U;wVkª Wuksa dh ceckjh djus ls ;g VwV tkrk gSA (D) nks gYds ukfHkd la;ksftr gksdj ,d Hkkjh ukfHkd cukrs gSa vkSj vU; laHko mRikn izkIr gksrs
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gSA
Page # 8
Chapter # 46
The Nucleus
Objective - II
1.
As the mass number A increases, which of the following quantities related toa nucleus do not change ? (A) mass (B) vloume (C*) density (D) binding energy tSls&tSls nzO;eku la[;k A c<+rh gS] ukfHkd ls lacaf/kr fuEu esa ls dkSulh jkf'k;k¡ ifjofrZr ugha gksrh gS (A) nzO;eku (B) vk;ru (C*) ?kuRo (D) ca/ku ÅtkZ
2.
The heavier nuclei tend to have larger N/Z ration becaues (A) a neutron is heavier than a proton (B) a neutron is an unstabel particle (C*) a neutron does not exert electric repulsion (D*) Coulomb forces have longer range compared to nuclear forces fdlh Hkkjh ukfHkd ds fy;s N/Z vuqikr vf/kd gksrk gS] D;ksfa d (A) izkVs kWu dh rqyuk esa U;wVkª Wu Hkkjh gksrk gSA (B) U;wVªkWu ,d vLFkk;h d.k gSA (C*) U;wVªkWu fo|qr izfrd"kZ.k ugha yxkrk gSA (D*) dwykWeh; cyksa dh ijkl ukfHkdh; cyksa dh rqyuk esa vf/kd gksrh gSA
3.
A free neutron decays to a proton but a free proton does not decay to a neutron. This is beacuse (A) neutron is a composite particle made of a proton and an electron whereas proton is fundamental particle (B) neutron is an uncharged particle whereas proton is a charged particle (C*) neutron has larger rest mass than the proton (D) weak forces can operate in a neutron but not in a proton.
,d Lora=k U;wVkª Wu dk izkVs kWu esa fo[k.Mu gks tkrk gS] fdUrq ,d Lora=k izkVs kWu dk fo[k.Mu U;wVkª Wu esa ugha gksrk gSA bldk dkj.k gS (A) U;wVªkWu ,d la;D q r d.k gS] tks ,d izkVs kWu vkSj ,d bysDVªkWu ls cuk gqvk gS] tcfd izkVs kWu ,d ewy d.k gSA (B) U;wVkª Wu ,d vukosf'kr d.k gS] tcfd izkVs kWu ,d vkosf'kr d.k gSA (C*) U;wVªkWu dk fojke nzO;eku] izkVs kWu ls vf/kd gSA (D) U;wVªkWu esa {kh.k cy yx ldrs gSa] fdUrq izkVs kWu esa ugha 4.
5.
6.
7.
8.
Consider a sample of a pure beta-active material (A) All the beta particles emitted have the same energy (B) The beta particles originally exist inside the nucleus and are ejected at the time of beta decay (C) The antineutrino emitted in a beta decay has zero mass and hence zero momentum. (D*) The active nucleus changes to one of its isobars after the beta decay ,d 'kq) chVk lfØ; inkFkZ ds uewus ij fopkj dhft;s (A) mRlftZr gksus okys leLr chVk d.kksa dh ÅtkZ ,d leku gksxhA (B) chVk d.kksa dk igys ls gh ukfHkd ds vUnj vfLrRo gksrk gS vkSj chVk&{k; ds le; budk mRltZu gks tkrk (C) chVk {k; esa mifLFkr izfr U;wfVªuksa dk nzO;ekuA (D*) chVk {k; esa mRlftZr izfr U;wfVªuksd a k nzO;eku 'kwU; gksrk gS] vr% budk laoxs Hkh 'kwU; gksrk gSA In which of the following decays the element does not change ? (A) -decay (B) +decay (C) --decay fuEu esa ls fdu fo[k.Muksa esa rRo ifjofrZr ugha gksrk gS (A) -{k; (B) + {k; (C) – {k; In which of the follwoing decaus the atomic number decreases ? (A*) -decay (B*) +decay (C) – decay fuEu esa ls fdu fo[k.Muksa esa ijek.kq Øekad de gksrk gS (A*) -{k; (B*) + {k; (C) – {k;
gSA
(D*) -decay (D*) – {k; (D) -decay (D) - {k;
Magnetic field does not cause deflection in (A) -rays (B) beta-plus rays
(C) beta-minus rays
(D*) gamma rays
pqEcdh; {ks=k ds dkj.k fdlesa fo{ksi.k ugha gksrk gS (A) -fdj.kksa esa (B) + {k;
(C) – {k;
(D*) - {k;
(C) beta-minus rays
(D*) gamma rays
Which of the follwoing are electromagnetic waves ? (A) -rays (B) beta-plus rays fuEu esa ls dkSulh fo|qr pqEcdh; rjaxsa gS -
manishkumarphysics.in
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Chapter # 46
The Nucleus
(A) -fdj.ksa
(B) /kukRed
chVk fdj.kksa esa
(C) _.kkRed
chVk fdj.kksa esa (D*) xkek fdj.kksa esa
9.
Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus because ? (A) a lithium nucleus is more tightly bound than a carbon nucleus (B) carbon nucleus is an unstable particle (C) it is not energetically favourable (D*) Coulomb repulsion does not allow the nuclei to come very close dejs ds rki ij yhfFk;e ok"i ij nks yhfFk;e ukfHkd la;ksftr gksdj dkcZu ukfHkd ugha cukrs gS]a D;ksfa d (A) dkcZu ukfHkd dh rqyuk esa yhfFk;e ukfHkd vf/kd n`<+rkiwod Z ca/kk jgrk gSA (B) dkcZu ukfHkd ,d vLFkk;h d.k gSA (C) ;g ÅtkZ ds vk/kkj ij laHko ugha gSA (D*) dwykeh; fod"kZ.k ukfHkdksa dks cgqr lehi ugha vkus nsrk gSA
10.
For nuclei with A > 100 (A) the binding energy of the nucleus decreases on an average as A increases (B*) the binding energy per nucleus decreases on an average as A increases (C*) if the nucleus breaks into two roughly equal parts, energy is released (D) if two nuclei fuse to form a bigger nucleus, energy is released. A > 100 okys ukfHkd ds fy;s (A) lk/kkj.k rkSj ij A esa o`f) ds lkFk ukfHkd dh ca/ku ÅtkZ de gksrh gSA (B*) lk/kkj.k rkSj ij A esa o`f) ds lkFk ukfHkd dh ca/ku ÅtkZ de c<+rh gSA (C*) ;fn ukfHkd yxHkx nks leku Hkkxksa esa VwVrk gS] rks ÅtkZ eqDr gksrh gSA (D) ;fn nks ukfHkd layf;r gksdj ,d cM+k ukfHkd cukrs gSa rks ÅtkZ eqDr gksrh gSA
WORKED OUT EXAMPLES 7.
Sol.
The half-life of 198Au is 2.7 days. Calculate (a) the decay constant, (b) the average-life and (c) the activity of 1.00 mg of 198Au. Take atomic weight of 198Au to be 198 g/mol. 198 Au dk v)Z&vk;qdy 2.7 fnu gSA crkb;s (a) {k; fu;rkad (b) vkSlr vk;q (c) 1.00 mg 198Au dh lfØ;rkA 198Au dk vk.kfod nzO;eku 198 g/mol yhft,A [HCV Chap46-WOE_Q7 ] Ans. (a) 2.9×10–6 (b) 3.9 days (c) 240 Ci. (a) The half-life and the decay constant are related as t1/2 = or,
=
ln 2 0.693
0.693 0.693 t1/ 2 = 2.7 days
0.693 = 2.7 x 24 x 3600 s = 2.9 x 10–6 s –1, (b) (c)
1 = 3.9 days The activity is A = N, Now, 198 g of 198Au has 6 x 1023 atoms. The number of atoms in 1.00 mg of 198Au is
The average-life is tav =
1.0 mg N = 6 x 1023 x 198 g = 3.03 x 1018 Thus,
A
= N = (2.9 x 10–6s –1) (3.03 x 1018) = 8.8 x 1012 disintegrations/s
8.8 x 1012 =
3.7 x 1010
Ci = 240 Ci.
manishkumarphysics.in
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Chapter # 46 The Nucleus 8. A radioactive sample has 6.0 × 1018 active nuclei at a certain instant. How many of these nuclei will still be in the same active state after two half-lives ? fdlh {k.k ,d jsfM;ks/kehZ inkFkZ esa 6.0 × 1018 lfØ; ukfHkd mifLFkr gSA bues ls fdrus ukfHkd nks v)Z vk;q ds ckn Hkh blh lfØ; voLFkk esa jgsaxs ? HCV_Ch-46_WOE_8 Sol. In one half-life the number of active nuclei reduces to half the original number. Thus, in two half-lives the 1 1
number is reduced to 2 2 of the original number. The number of remaining active nuclei is, therefore,
6.0 × 1018
1 × 2
1 2
= 1.5 × 1018
EXERCISE Mass of proton m p = 1.007276 u , Mass of 11H atom = 1.007825 u , Mass of Neutron m n = 1.008665 u , Mass of electron = 0.0005486 u 511 kv V/c 2 , 1u = 931 MeV/c 2
izksVkWu dk nzO;eku m p = 1.007276 u , 11H ijek.kq dk nzO;eku = 1.007825 u , U;wVªkWu dk nzO;eku m n = 1.008665 u , bysDVªkWu dk nzO;eku = 0.0005486 u 511 kv V/c2 , 1u = 931 MeV/c2 1.
Assume that the mass of a nucleus is approximately given by M = Amp where A is the mass number. Estimate the density of matter in k/g m 3 inside a nucleus .What is the specific gravity of numlear matter ? eku yhft;s fd fdlh ukfHkd ds nzO;eku dk lfUudVre M = Am p }kjk O;Dr fd;k tkrk gS] tgk¡ A nzO;eku la[;k gSA
fdlh ukfHkd ds vUnj ds ?kuRo dk vuqeku fdxzk@eh3 esa yxkb;sA ukfHkdh; inkFkZ dk vkisf{kd ?kuRo fdruk gSA Ans:
3 × 10 17 kg/m 3 , 3 × 10 14
2.
A neutron star has a density equal to that of the nuclear matter. Assuming the star to be spherical fine the radius of a neutron star whose mass is 4.0 × 10 30 kg (twice the mass of the sun)
fdlh U;wVkª Wu rkjs dk ?kuRo ukfHkdh; inkFkZ ds cjkcj gksrk gSA eku yhft;s fd rkjk xksykdkj gSA bl U;wVkª uW rkjs dh f=kT;k Kkr dhft;s] ftldk nzO;eku 4.0 × 10 30 fdxzk (lw;Z ds nzO;eku dk nqxuk) gSA Ans:
15 km
3.
Calculate the mass of an particle ts binding energy is 28.2 Me.V ,d d.k ds nzO;eku dh x.kuk dhft;sA bldh ca/ku ÅtkZ 28.2 Me.V gSA 4.0016 u
Ans: 4.
How much energy is released in the following reaction ? Li + P + 7 4 Atomic mass of Li = 7.0160 u and that of He = 4.0026u.
fuEufyf[kr vfHkfØ;k esa fdruh ÅtkZ eqDr gksxh\ Ans:
Li + P + 7 Li dk ijekf.od nzO;eku = 7.0160 u vkSj 17.34 Me V
5.
Find the binding energy per nucleon of 197 79 Au if its atomic mass is 196.96 u.
;fn
197 79 Au
bldh izfr U;wfDyvkWu ca/ku ÅtkZ Kkr dhft;sA He = 4.0026u 4
dk ijek.kq Hkkj 196.96 u gS] rks bldh izfr U;wfDyvkWu ca/ku ÅtkZ Kkr dhft;sA
Ans:
7.94 MeV
6.
(a) Calculate the energy released if 238 U emits an a-particle. (b) Calculate the energy to be supplied to 238 U if two protons and two neutrons are to be emitted one by one The atomic masses of 238 U , 238 Th and 4He are 238.0508 u , 234 .04363 u and 4.00260 u respectively (a) ;fn 238 U ,d -d.k mRlftZr djrk gS rks eqDr ÅtkZ dh x.kuk dhft;sA (b) ,d&,d djds nks izksVkWu mRlftZr gks rks 238 U dks nh x;h ÅtkZ dh x.kuk dhft;sA 238 U , 238 Th ,oa 4He ds ijekf.od Hkkj Øe'k% 238.0508 u , 234.04363 u ,oa 4.00260 u gSAa (a) 4.255 Me V (b) 24.03 Me V
Ans:
manishkumarphysics.in
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Chapter # 46 7.
Ans: 8.
The Nucleus
Find the energy liberated in the reaction fuEu vfHkfØ;k esa eqDr 238 209 Ra Pb + 14C. The atomic masses needed are as follows : vko';d ijekf.od 238 209 14 Ra Pb C 223.018 u 208.981u 14.003 u 31.65 Me V
ÅtkZ Kkr dhft;s nzO;eku fuEu gS %
Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is E = (Mz = 1,N + MH – Mz,N)c 2 Where Mz, N = mass of an atom with Z protons and N neutrons in the nucleus and MH mass of a hydrogen atom This energy is known as proton separation energy O;Dr dhft;s fd Z izkVs kWuksa ,oa N U;wVkª Wuksa okys ukfHkd ls ,d izkVs kWu vyx djus ds fy;s vko';d U;wure ÅtkZ fuEu gS% E = (Mz = 1,N + MH – Mz,N)c 2 tgk¡ Mz, N = ukfHkd esa Z izkVs kWu ,oa N U;wVkª Wu okys ijek.kq dk nzO;eku vkSj MH = gkbMªkt s u ijek.kq dk nzO;ekuA ;g ÅtkZ]
^^izkVs kWu i`FkDdhdj.k ÅtkZ** dgykrh gSA Ans : 9.
Ans: 10.
Ans: 11.
Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons in terms of the masses Mz,N, M zN–1 and the mass of the neutron. Z izkVs kWu ,oa N U;wVªkWu okys ukfHkd ls ,d U;wVkª Wu dks i`Fkd djus ds fy;s vko';d U;wure ÅtkZ dh x.kuk Mz, N , Mz , N–1 vkSj U;wVªkWu ds nzO;eku ds inksa esa dhft;sA (Mz, N– 1 m n – Mz, N)c 2 P beta decays to 32S Find the sum of the energy of the antineutrino and the kinetic energy of the particle Neglect the recoil of the daughter nucleus. Atomic mass of 32 P = 31.974 u and that of 32S = 31.972 u. 32 P dk chVk {k; 32S esa gksrk gSA izfr U;wfVªuksa ¼izfr U;wfVªuks½a rFkk -d.k dh xfrt ÅtkZ dk ;ksx Kkr dhft;sA mRikn ukfHkd dk izfr{ksi xkS.k dj nhft;sA 32P dk ijek.kq Hkkj = 31.974 u vkSj 32S dk ijek.kq Hkkj 31.972 u gSA 1.86 Me V 32
A free neutron beta-decays to a proton with a half life of 14 minutes (a) What is the decay constant ?(b) Find the energy liberated in the process. ,d Lora=k U;wVkª Wu dk 14 fefuV v)Z&vk;q ds lkFk izkVs kWu esa chVk {k; gksrk gS : (a) {k; fu;rkad fdruk gS\ (b) bl U;wfVªuksa
dk laosx fdxzk&eh@ls esa fdruk gS\ QksVkWu ij ykxw gksus okyk lw=k iz;qDr dhft;sA –4
(a) 8.25 × 10
12.
Complete the following decay schemes. fuEu (a)
226 88 Ra
s
–1
Ans:
(b) 782 Ke V
(B)
fo[k.Mu vfHkfØ;kvksa dks iwjk dhft;s % 19 8 O
19 9 F
25 12 Mg
(C)
25 13 Al
Ans:
(a)
222 86 Rn
13.
In the decay 64 Cu 64 Ni + e + v, the maximum kinetic energy carried by the positron is found to be 0.650 Me V (a) What is the energy of the neutron which was emitted together with a positron of kinetic energy 0.150 Me V ? (b) what is the momentum of this neutrino in kg - m /s ?Use the formula applicable to photon fo[k.Mu fØ;k 64 Cu 64 Ni + e + v, esa ikWftVªkWu dh vf/kdre xfrt ÅtkZ 0.650 MeV gSA (a) 0.150 Me V ÅtkZ okys ikWftVªkWu ds lkFk mRlftZr gksus okys U;wfVªuksa dh ÅtkZ fdruh gS\ (b) bl U;wfVªuksa dk laosx fdxzk&eh@ls esa fdruk
(B) e v (c) e + + v
gS\ QksVkWu ij ykxw gksus okyk lw=k iz;D q r dhft;sA Ans: 14.
(a) 500 Ke V (b) 2.67 × 10 – 22 kg/ – m/s Potassium -40 can decay in three modes It can decay by – emission + emission or electron capture (a) Write the equations showing the end products .(b) Find the Q- values in each of the three cases. Atomic masses of
40 40 18 Ar , 19 K
and
40 20 Ca
are 39.9624 u , 39.9640u and 39.9626 u respectively..
iksVfs 'k;e-40 dk fo[k.Mu rhu fo/kkvksa esa gks ldrk gSA bldk fo[k.Mu – mRltZu ls] + mRltZu ls ;k bysDVªkWu izxzg.k ls gks ldrk gSA (a) vafre mRiknksa dks n'kkZus okyh lehdj.kksa dks fyf[k;sA (b) rhuksa izfØ;kvksa esa ls izR;sd ds fy;s Q-eku
manishkumarphysics.in
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Chapter # 46
The Nucleus
Kkr dhft;sA and Ans:
(a)
40 40 19 K 20Ca
40 40 18 Ar , 19 K
,oa
40 20 Ca
ds ijek.kq Hkkj Øe'k% 39.9624 u , 39.9640u ,oa 39.9626 u gSAa
40 40 40 40 e v , 10 K 18 Ar e v, 19 K e 18 Ar v
(b) 1.3034 MeV, 0.4676 MeV, 1.490MeV 15.
Lithium (Z = 3) has two stable isotopes 6Li and 7Li .When neutrons are bombarded on lithium sample electrons and -particles are ejected. Write down the nuclear processes taking place. yhfFk;e (Z = 3) ds nks LFkk;h leLFkkfud 6Li o 7Li gksrs gSaA tc yhfFk;e ds uewus ij U;wVkª Wu dh ceckjh dh tkrh gS] rks bysDVªkWu vkSj -d.k mRlftZr gksrs gSaA blds fy;s gksus okyh ukfHkdh; vfHkfØ;k fyf[k;sA
Ans:
6 3 Li
16.
The masses of 11C and 11B are respectively 11.0114 u and 11.0093 u Find the maximum energy a positron can have in the decay of 11 C to 11B. 11 C vkSj 11B ds nzO;eku Øe'k% 11.0114 u o 11.0093 u gSA 11 C ls 11B esa gksus okys {k; esa mRlftZr ikWftVªkWu dh
n 73 Li n 73 Li 83 Li 84 Bi e v , 84 Bi 24He 24He
vf/kdre ÅtkZ Kkr dhft;sA Ans :
933.6 Ke V
17.
238
The emits an alpha particle to reduce to 224 Ra .Calculate the kinetic energy of the alpha particle emitted in the following decay : [HCV_Chp_46_Ex. 17] 238 224 Th ,d ,YQk&d.k mRlftZr djds Ra esa ifjofrZr gks tkrk gSA fuEu fo[k.Mu esa mRlftZr ,YQk d.k dh xfrt ÅtkZ
dh x.kuk dhft;sA Ans :
5.304 Me V 228 Th 224 Ra + 224 Ra 224 Ra + ( 217 ke V). Atomic mass of 238Th is 228.028726 u That of 224 Ra is 224.0200196 u and that of 238
18.
Th kdk
ijek.kq Hkkj 228.028726 u gS] vkSj
224
Ra dk 224.0200196 u rFkk 24He
4 2He
is 4.00260 u
dk 4.00260 u gSA
Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme
fuEu {k; fØ;k esa mRlftZr chVk d.k dh vf/kdre xfrt ÅtkZ dh x.kuk dhft;sA N 12C * + e + v C * 12C + (4.43 MeV.) The atomic mass of 12N is 12.018613 u . 11.88 MeV 12 12
Ans: 19.
12
N dk
ijek.kq Hkkj 12.018613 u gSA
The decay constant of 197 (electron capture to 197 ) is 1.8 × 10 – 4 s – 1 (a) .What is the half life ? 79 Hg 79 Hg (b) What is the average life ? (c) How much time will it take to convert 25 % of this isotope of mercury into gold ?
dk {k; fu;rkad ( 197 ds bysDVªkWu f'kdats çfØ;k es) 1.8 × 10 – 4 s – 1 gSA (a) bldh v)Z vk;q fdruh gS ? (b) 79 Hg bldh vkSlr vk;q D;k gS ? (c) edZjh ds bl leLFkkfud ds 25% dks ] lksus esa cnyus esa fdruk le; yxsxk ? 197 79 Hg
HCV_Ch-46_Ex._19
Ans: 20.
Ans: 21.
Ans: 22.
6930 10000 seconds = 64 min (b) seconds = 92 min (c) 92 n 4 min.= 1600s (a) 1.8 1.8 3
The half-life of 198Au is 2.7 days. (a) Find the activity of a sample containing 1.00 µg of 198 Au(b) What will be the activity after 7 days ? Take the atomic weight of 198 Au to be 198 g/mol. 198 Au dh v)Z vk;q 2.7 fnu gSA (a) ml uewus dh lfØ;rk Kkr dhft, ftlesa 198 Au ds 1.00 µg gSA (b) 7 fnu i'pkr~ lfØ;rk D;k gskxh ? 198 Au dk vk.kfod nzO;eku 198 g/mol yhft,A (a) 0.244 Ci (b) 0.040 HCV_Ch-46_Ex_20 Radioactive 131 has a half-life of 8.0 days A sample containing 131 has activity 20 µ Ci at t = 0 (a) What is its activity at t = 4.0 days ? (b) What is its decays constant at t = 4.0 jsfM;ks&lfØ; 131 dh v)Z&vk;q 8.0 fnu gSA ,d uewuk ftlesa 131 fo|eku gS] dh lfØ;rk t = 0 ij 20 µ Ci gSA (a) t = 4.0 fnu ij bldh lfØ;rk fdruh gksxh\ (b) t = 4.0 ij bldk {k; fu;rkad fdruk gksxkA (a) 14 µ Ci (b) 1.4 × 10 – 6 s – 1 The decay constant of 238U is 4.9 × 10 – 18 s – 1 (a) What is the average life of 238U ?(b) What is the halflife of 238U ? (c) By what factor does the activity of a 238 U sample decrease in 9 × 109 years? manishkumarphysics.in
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Chapter # 46
The Nucleus
U dk {k; fu;rkad 4.9 × 10 izfr ls- gSA (a) What is the average life of 238U ?(b) What is the half-life of U ? (c) By what factor does the activity of a 238 U sample decrease in 9 × 109 years? k 6.49 × 109 y (b) 4.5 × 10 9 y (c) 4 238
– 18
238
Ans: 23.
Ans: 24.
Ans: 25.
Ans: 26.
A certain sample of a radioactive material decays at the rate of 500 per second at a certain time The count rate falls to 200 per second after 50 minutes (a) What is the dacey constant of the sample ?(b) What is its half-life ? ,d jsfM;ks lfØ; inkFkZdk dksbZ uewuk fdlh fuf'pr le; ij 500 izfr lsd.M dh nj ls fo[kf.Mr gks jgk gSA 50 fefuV i'pkr~ x.ku nj 200 izfr lsd.M jg tkrh gSA (a) uewus dh {k;&nj fdruh gSA (b) bldh v)Z&vk;q fdruh gS\ 3.05 × 10 – 4 s (b) 38 min The count rate from a radioactive sample falls from 4.0 × 10 6 per seconds to 1.0 × 10 6 per second in 20 hours what will be the count rate 100 hours after the beginning ? ,d jsfM;ks lfØ; uewus esa 20 ?kaVksa esa x.kuk nj 4.0 × 106 izfr lsd.M ls de gksdj 1.0 × 106 izfr lsd.M jg tkrh gSA izkjEHk ls 100 ?kaVksa ds i'pkr~ x.kuk nj fdruh gksxh\ 3.9 × 10 3 per second The half-life of 226Ra is 1602 y .Calculate the activity of 0.1 g of RaCl2 in which all the radium in the from of 220Ra Taken atomic weight of Ra to be 226 g/mol and that of Cl to be 35.5 g/mol 226 Ra dh v)Z&vk;q 1602 o"kZ gSA 0.1 xzkz e RaCl2 dh lfØ;rk dh x.kuk dhft;s] ftlesa lEiw.kZ jsfM;e 226Ra voLFkk dk gh gSA Ra dk ijek.kq Hkkj 226 xzke@eksy rFkk Cl dk 35.5 xzke@eksy eku yhft;sA 2.8 × 10 9 disintergrations/s The half-life of a radioisotope is 10 h .Find the total number of disintegrations in the tenth hour measured from a time when the activity was 1 Ci. ,d jsfM;ks&leLFkkfud dh v)Z&vk;q 10 ?kaVs gSA tc bldh lfØ;rk 1 Ci Fkh] rc ls nlosa ?kaVs esa ekih x;h] dqy fo[k.Muksa
dh la[;k Kkr dhft;sA Ans:
6.91 × 10 13
27.
The selling rate of a radioactive isotope is decided by its activity what will be the second hand rate of a one month old 32 P(t1/2 = 14.3 days ) source if it was originally purchased for 800 rupees ? jsfM;ks lfØ; leLFkkfudksa dk foØ; ewY; budh lfØ;rk ds vk/kkj ij r; fd;k tkrk gSA ,d eghus iqjkus 32P(t1/2 = 14.3 fnu) lzksr dh lsd.M&gs.M fdruh gksxh] ;fn bldks izFke ckj 800 :i;s ls Ø; fd;k x;k Fkk\ 187 rupees
Ans: 28.
Co dacys to 57 Fe by - emission The resulting 57 Fe is in its excited state and comes to the ground state by emitting state and comes to the ground state by emitting -rays emission is 10 – s A sample of 57 Co gives 5.0 × 109 gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5 × 10 9 per second ? 57 Co dk + mRltZu ls 57Fe esa fo[k.Mu gksrk gSA ifj.kkeh 57Fe mÙksftr djds ewy&voLFkk esa vk tkrk gS rFkk -fdj.kksa dk mRltZu djds ewy&voLFkk esa vk tkrk gSA + - {k; dh v)Z vk;q 270 fnu vkSj -{k; dh 10–8 lsd.M gSA 57Co dk ,d izfrn'kZ izfr lsd.M 5.0 × 109 xkek fdj.ksa iznku dj jgk gSA xkek fdj.kksa dh mRltZu nj 2.5 × 109 izfr lsd.M rd 57
de gksus ls iwoZ fdruk le; O;rhr gks pqdk gksxk\ Ans:
270 days
29.
Carbon (Z = 6 ) with mass number 11 decays to boron (Z = 5 ) (a) Is it a decay or b decay ? (b) The half-life of the decay scheme is 20.3 minutes .How much time will elapse before a mixture of 90% carbon 11 and 10% boron -11 (by the number of atoms ) converts itself into a mixture of 10% carbon-11 and 90% boron-11? – dkcZu (Z = 6 ) ftldh nzO;eku la[;k 11 gS] cksjksu (Z = 5) esa fo[kf.Mr gksrk gSA (a) D;k ;g {k; gS vFkok - {k;\ (b) fo[k.Mu izfØ;k dh v)Z&vk;q 20.3 fefuV gSA 90% dkcZu -11 o 10% cksjksu -11 ds feJ.k ¼ijek.kqvksa dh la[;k ds
vk/kkj ij½ dks ifjofrZr gksus esa fdruk le; yxsxk\ Ans :
(a) + (b) 64 min
30.
Ans :
4 × 1023 tritium atoms are contained in a vessel. The half-life of decay of tritium nuclei is 12.3 Find (a) the activity of the sample ,(b) the number of decays in the next 10 hours (c) the number of decays in the next 6.15 y. ,d ik=k esa VªhfV;e ds 4 × 1023 ijek.kq gSA VªhfV;e ukfHkd ds fo[k.Mu dh v)Z&vk;q 12.3 o"kZ Kkr dhft;s % (a) izfrn'kZ ¼uewus½ dh lfØ;rk (b) vxys 10 ?kaVksa esa fo[k.Muksa dh la[;k (c) vxys 6.15 o"kks± esa fo[k.Muksa dh la[;kA (a) 7.146 × 10 14 disintergrations/s
31.
A point source emitting alpha particles is placed at a distance of 1m from a counter which records any manishkumarphysics.in
Page # 14
Chapter # 46 The Nucleus alpha particle falling on its 1cm 2 window .If the source contains 6.0 × 1016 active nuclei and the counter records a rate of 50000 count /second find the decay constant Assume that the source emits alpha particles fall nearly in all directions and the alpha particles fall nearly normally on the window. 2 vYQk d.kksa dks mRlftZr djus okyk fcUnq lzkrs ,d dkm.Vj ¼x.kd½ ls 1eh- nwj j[kk gqvk gS] tks bldh 1lseh fo.Mksa ij vkifrr fdlh Hkh vYQk&d.k dks fjdkMZ dj ysrk gSA ;fn lzksr esa 6.0 × 1016 lfØ; ukfHkd gSa vkSj dkm.Vj 50000
[email protected] dh nj ls fjdkMZ dj jgk gS] rks {k; fu;rkad Kkr dhft;sA ;g eku yhft;s fd lzkrs leLr fn'kkvksa esa ,d leku :i ls vYQk d.k mRlftZr dj jgk gS rFkk vYQk d.k f[kM+dh ij yxHkx yEcor~ vkifrr gks jgs gSAa
Ans:
1.05 × 10
32.
238
–7
s
–1
U decays to 20Pb with a half-life of 4.47 × 109 y. This happens in a number of steps Can you justify a single half-life for this chain of processes? A sample of rock is found to contain 2.00 mg of 238U and 0.600 mg of 20 Pb Assuming that all the lead has come from uranium find the life of the rock. 238 U dk 4.47 × 109 o"kZ v)Z&vk;q ds lkFk 206Pb esa fo[k.Mu gksrk gSA ;g dbZ inksa esa ?kfVr gksus okyh izfØ;k gSA D;k vki bl Ja[` kyk izfØ;k ds fy;s] ,d vdsyh v)Z&vk;q fu/kkZfjr dj ldrs gSa\ fdlh pV~Vku ds uewus esa 2.00 fexzk 238U o 0.600 mg 20Pb ik;k x;k A eku yhft;s fd lhls dh lEiw.kZ ek=kk ;wjfs u;e ls gh izkIr gqbZ gS] rks pV~Vku dh vk;q Kkr
dhft;sA Ans:
1.92 × 10 9 y
33.
When charcoal is prepared from a living tree, it shows a disintergration rate of 15 .3 disintergrations of 14 C per gram per minute A sample from an ancient piece of charcoal show 14C activity to be 12.3 disintergrations per gram per minute How old is this sample ? Half-life of 14C is 5730y. tc thfor o`{k ls dks;yk cuk;k tkrk gS] rks ;g 14C ds fo[k.Mu dh nj 15.3 fo[k.Mu izfrxzke izfr fefuV n'kkZrk gSA dks;ys dk ,d izkphu VqdM+k 14C dh lfØ;rk 12.3 fo[k.Mu izfr xzke izfr fefuV n'kkZ jgk gSA ;g uewuk fdruk iqjkuk gSA 14C dh v)Z vk;q 5730 o"kZ gSA [HCV_Chp_46_Ex. 33] 1800 y
Ans : 34.
Natural water contains a small amount of tritium (13 H) This isotope beta decays with a half-life of 12.5 years A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things he finds a sealed bottle of whisky On return he analyses the whisky and finds that it contains only 1.5 per cent of the (13 H) radioactivity as compared to a recently purchased bottle marked 8 years old’ Estimate the time of that unsuccessful attempt.
izkÑfrd ty esa VªhfV;e (13 H) dh dqN ek=kk fo|eku jgrh gSA bl leLFkkfud dk 12.5 o"kZ v)Z&vk;q ds lkFk chVk&{k; gksrk gSA ,d ioZrkjksgh dks ,d dfBu p<+kbZ okyh pksVh dh vksj p<+rs gq, fdlh iwooZ rhZ vlQy iz;kl ds vo'ks"k feyrs gSAa ftlesa vU; oLrqvksa ds lkFk fOgLdh dh lhy can cksry Hkh feyrh gSA okil ykSVus ij og fOgLdh dk fo'ys"k.k djrk gS] vkSj ;g ikrk gS fd blesa vHkh [kjhnh gqbZ "8 o"kZ iqjkuh" cksry dh rqyuk esa (13 H) dh lfØ;rk dsoy 1.5% gh gSA ml vlQy iz;kl ds dky dk vuqeku yxkb;sA Ans:
about 83 years go
35.
The count rate of nuclear radiation coming from a radioactive sample containing 128 varies with time as follows . Time t ( minute ): 0 25 50 75 100 Count rate R (109 s – 1) : 30 16 8.0 3.8 2.0 (a) Plot in (Ro/R ) against t (b) From the slope of the best straight line through the points find the decays constant (c) Calculate the half-life t 1/2. ,d jsfM;kslfØ; uewus esa 128 fo|eku gS] blls vkus okys ukfHkdh; fofdj.k dh x.kuk nj le; ds lkFk fuEukuqlkj ifjofrZr gksrh gS : le; t (fefuV): 0 25 50 75 100 9 –1 x.ku nj R (10 s ) : 30 16 8.0 3.8 2.0 (a) le; t ds lkis{k In (R0/R) dk xzkQ [khafp;sA (b) fcUnqvksa ls xqtjusokyh lokZf/kd lh/kh ljy js[kk ds >qdko ds vk/kkj ij {k; fu;rkad Kkr dhft;sA (c) v)Z&vk;q t1/2 dh x.kuk dhft;sA (b) 0.028 min – 1 approx (c) 25 min approx
Ans: 36.
The half-life of 40K is 1.30 × 10 9 y A sample of 1.00g of pure KCI gives 160 counts/s Calculate the relative abundance of 40K (fraction of 40K present ) in natural potassium. [HCV_Chp_46_Ex. 36] 40 K dh v)Z&vk;q 1.30 × 109 o"kZ gSA KCl dk 1.00 xzke dk uewuk 160
[email protected] dj jgk gSA izkÑfrd iksVfs 'k;e esa 40K dh miyC/krk (40K dk izfr'kr Hkkx½ dh x.kuk dhft;sA manishkumarphysics.in
Page # 15
Chapter # 46 Ans: 0.12% 37.
197 80 Hg
The Nucleus
decays to
197 79 Au
through electron capture with a decay constant of 0.257 per day (a) what other
particle or particles are emitted in the decay ? (b) Assume that the electrons is captured from the K 7 –1/2 v = a (Z – b ) with a = 4.95 × 10 s and b = 1 to find the wavelength of the ka Xray emitted following the electrons capture.
shell Use Mosely’s law
bysDVªkWu izxzg.k }kjk 197 79 Au esa fo[kf.Mr gksrk gS] ftldk {k; fu;rkad 0.257 izfr fnu gSA (a) bl fo[k.Mu esa vU; dkSulk ;k dkSuls d.k vkSj mRlftZr gksrs gSAa (b) eku yhft;s fd bysDVªkuW dk izxgz .k K-d{kd ls gksrk gSA ekslys ds fu;e v = a (Z – b ), a = 4.95 × 107 s–1/2 o b = 1 dh lgk;rk ls K X-fdj.kksa dh rjaxnS/;Z Kkr dhft;s tks bysDVªkWu izxzg.k ds dkj.k mRlftZr gksrh gSA 197 80 Hg
Ans:
(a) neutrino (b) 20 pm
38.
A radioactive isotope is being produced at a constant rate dN/dt = R in an experiment .The isotope has a half-life t1/2. Show that after a time t >> t1/2, the value of this constant. ,d iz;ksx esa ,d jsfM;ks lfØ; leLFkkfud dk mRiknu fu;r nj dN/dt = R ls gks jgk gSA leLFkkfud dh v)Z&vk;q t1/2 gSA O;Dr dhft;s fd le; t >> t1/2 ds i'pkr~ lfØ; ukfHkdksa dh la[;k fu;r gks tk;sxhA fu;rkad dk eku Kkr dhft;sA
Ans: 39.
Ans : 40.
Ans : 41.
Rt1/ 2 0693
Consider the situation of the previous problem Suppose the production of the radioactive isotope starts at t = 0 Find the number of active nucleai at time t. fiNys iz'u esa of.kZr ifjfLFkfr ij fopkj dhft;sA eku yhft;s fd jsfM;ks lfØ; leLFkkfud dk mRiknu t = 0 ij izkjEHk gksrk gSA le; t ij lfØ; ukfHkdksa dh la[;k Kkr dhft;sA R (1 e t )
In an agricultural experiment ,a solution containing 1 mole a of radioactive material (t 1/2 = 14.3 days ) was injected into the roots of a plant .The plant was allowed 70 hours to settle down and then activity was measured in its fruit If the activity measured was 1 µ Ci what per cent of activity is transmitted from the root to the fruit in steady state ? Ñf"k lEcfU/kr ,d iz;ksx es]a ,d eksy jsfM;ks lfØ; inkFkZ (t1/2 = 14.3 fnu) dk foy;u ikS/ks dh tM+kas esa batD s 'ku esa Mkyk x;kA ikS/ks dks 70 ?kaVksa rd lqfLFkj gksus ds fy;s NksMk+ x;k rRi'pkr~ blds Qyksa esa lfØ;rk ekih x;ha ;fn ekih x;h lfØ;rk 1 µ Ci gSa] rks LFkk;h fLFkfr esa tM+ ls Qy rd fdrus izfr'kr lfØ;rk LFkkukarfjr gqbZ\ 1.26 × 10 – 11 % A vessel of volume 125 cm 3 contains tritium (3H t1/2 = 12.3 y ) at 500 kpa and 300 K Calculate the activity of the gas 125 lseh3 vk;ru okys ik=k esa 500 kPa o 300 K VªhfV;e (3H t1/2 = 12.3 o"kZ) \ Hkjk gqvk gSA xSl dh lfØ;rk dh x.kuk
dhft;sA Ans :
724 Ci
42.
212 83 Bi
can disintegrate either by emitting an -particle or emitting a -particle (a) Write the two equations
showing the products of the decays (b) The probabilities of disintergration by a decays are in the ratio 7/13 The overall half life of 212Ni is one hour If 1 g of pure 212Bi is taken at 12.00 noon ,what will be the composition of this sample at 1 p.m. the same day?
dk fo[k.Mu -d.k vFkok –-d.k ds mRltZu ls gks ldrk gSA (a) fo[k.Mu mRiknksa dks iznf'kZr djus okyh nks lehdj.ksa fyf[k;sA (b) rFkk - fo[k.Muksa dh izkf;drkvksa dk vuqikr 7/13 gSA dqy feykdkj 212Bi dh v)Z&vk;q ,d ?kaVk gSA ;fn nksigj 12.00 cts 1 xzke 'kq) 212Bi fy;k tk;s rks mlh fnu 1.00 pm, ij bl uewus esa fdl&fdl rRo dh fdruh ek=kk gksxh\ 212 83 Bi
212 208 212 212 83 Bi 81 TI , 83 Bi 84
Bi 212 84 Po e v
Ans :
(a)
43.
A sample contains a mixture of 108 Ag and 110Ag isotopes each having an activity of 8.0 × 10 8 disintergrations per second .110 Ag is known to have larger half-life than 108Ag. The activity A is measured as a function of time and the following data are obtained. (a) Plot In (A/Ao) versus time (b) See that for large values of time the plot is nearly linear Deduce the manishkumarphysics.in
Page # 16
Chapter # 46 The Nucleus half-life of 110Ag from this portion of the plot (c)Use the half-life of 110 Ag to calculate the activity corresponding to 108 Ag in the first 50 S (d) plot in (A/Ao) versus time for 108 Ag for the first 50 s. (e) Find the half-life of 108 Ag
Time (s) 20 40 60 80 100
Activity (A) (108 Activity (A) (108 Time (s) disintegrations's) disintegrations/s) 11.799 9.168 7.4492 6.2684 5.4115
200 300 400 500
3.0828 1.8899 1.1671 0.7212
,d uewus esa 108Ag vkSj 110 Ag leLFkkfudksa dks feJ.k gS] izR;sd dh lfØ;rk 8.0 × 108 fo[k.Mu izfr lsd.M gSA 108Ag ls vf/kd gSA le; ds lkFk lfØ;rk dk ekiu djus ij fuEu vkadM+s izkIr gq, : (a) le; ds lkis{k In (A/Ao) dk xzkQ [khafp;sA (b) ;g nsf[k;s fd le; ds vf/kd ekuksa ds fy;s xzkQ yxHkx jSf[kd gS] xzkQ ds bl Hkkx ls 110Ag dh v)Z&vk;q Kkr dhft;sA (c) 110Ag dh v)Z&vk;q iz;D q r djds izFke 50 lsd.Mksa esa 108Ag ds fy;s 108 le; ds lkis{k (A/Ao) dk xzkQ [khafp,A (e) Ag dh v)Z&vk;q Kkr dhft;sA A½ A½ l fØ; r k¼ l fØ; r k¼ l e; ¼ l s-½ 108 l e; ¼ ls -½ 108 ¼ fo[ k.Mu@l s ½ ¼ fo[ k.Mu@l s ½ 20 40 60 80 100
11.799 9.168 7.4492 6.2684 5.4115 110
200 300 400 500
3.0828 1.8899 1.1671 0.7212
Ag 24.4 s and of
108
Ans :
the half life of
Ag = 144s
44.
A human body excretes (removes by waste discharge sweating etc.) certain materials by a low similar to radioactivity If technetium is injected in some form in a human body , the excretes half the amount in 24 hours A patient is given an injection containing 99 Tc this isotope is radioactive with a half -life of 76 hours The activity from the body just after the injection is 6 µ Ci How much time will elapse before the activity falls to 3 µ Ci ?
ekuo 'kjhj ls dqN inkFkks± dk fu"dklu ¼ey&ew=k rFkk ilhus }kjk½ jsfM;ks lfØ;rk ds tSls fu;ekuqlkj gh fd;k tkrk gSA;fn VsfDuf'k;e dh fdlh voLFkk dks batD s 'ku }kjk ekuo 'kjhj esa Mky fn;k tk;s 24 ?kaVksa esa rks 'kjhj bldh vk/kh ek=kk ckgj fudky nsrk gSA ,d ejht dks 99Tc dk batD s 'ku fn;k x;kA ;g leLFkkfud jsfM;ks lfØ; gS] ftldh v)Z vk;q 6 ?kaVs gSA batsD'ku ds rqjar i'pkr~ 'kjhj esa lfØ;rk 6 µ Ci gSA lfØ;rk dk eku 3 µ Ci rd de gksus esa fdruk le; yxsxk ? Ans :
4.8 hours
45.
A charged capacitor of capacitance C is discharged through a resistance R A radioactive sample decays with an average life Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time. C /kkfjrk dk ,d vkosf'kr la/kkfj=k] izfrjks/k R }kjk folftZr fd;k tkrk gSA ,d jsfM;kslfØ; uewuk vkSlr&vk;q ds lkFk fo[kf.Mr gksrk gSA R dk og eku Kkr dhft;s] ftlds fy;s fdlh le; ds lkFk la/kkfj=k esa lafpr fLFkj&oS|rq ÅtkZ rFkk
jsfM;kslfØ; uewus dh lfØ;rk vuqikr fuf'pr jgsA Ans :
2/C
46.
Radioactive isotopes are produced in a nuclear physics experiment at a constant rate dN/dt = An inductor of inductance 100 mH a resistor of resistance 100 and a battery are connected to from a series circuit .The circuit is switched on at the instant the production of radioactive isotope starts It is found that i/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t Find the half-life of the isotope. ukfHkdh;&HkkSfrdh ds iz;ksx esa jsfM;ks&lfØ; leLFkkfud dk mRiknu fu;r nj dN/dt = R ls gksrk gSA 100 mH dk izjs dRo] 100 izfrjks/k o ,d cSVjh dks Js.kh ifjiFk esa tksM+k x;k gSA ifjiFk dk fLop ml le; ij pkyw fd;k tkrk gS] tc jsfM;ks&lfØ;leLFkkfud dk mRiknu izkjEHk gksrk gSA ;g ik;k x;k fd le; ds lkFk i/N fu;r jgrk gS] tgk¡ i le; t ij ifjiFk esa /kkjk gS] vkSj N le; t ij lfØ; ukfHkdksa dh la[;k gSA leLFkkfud dh v)Z&vk;q Kkr dhft;sA 6.93 × 10 4 K
Ans: 47.
Calculate the energy released by 1 g of natural uranium assuming 200 Me V is released in each fission event and that the fissioable 235 U has an abundance of 0.7 % by weight in natural uranium 1xzke izkÑfrd ;wjfs u;e }kjk eqDr ÅtkZ dh x.kuk ;g ekurs gq, dhft;s fd izR;sd fo[k.Mu esa 200 MeV ÅtkZ eqDr gksrh manishkumarphysics.in
Page # 17
Chapter # 46
The Nucleus
gS rFkk izkÑfrd ;wjsfu;e esa fo[k.Mu ;ksX; leLFkkfud
235
U dk 0.7% Hkkj
fo|eku jgrk gSA
8
Ans:
5.7 × 10 J
48.
A uranium reactor dipoles thermal energy at a rate of 300 MW Calculate the amount of 235 U being consumed every second . Average energy released per fission is 200 Me V. ,d ;wjsfu;e&fj,DVj 300 MW dh nj ls Å"ek&ÅtkZ mRikfnr dj jgk gSA 235U dh izfr lsd.M iz;D q r ek=kk dh x.kuk dhft;sA izfr fo[k.Mu vkSlru 200 MeV ÅtkZ eqDr gksrh gSA 3.7 mg
Ans: 49.
A town has a population of 1 million .The average electric power needed per person is 300 W. A reactor is to be designed to supply power is converted into electric with which thermal power is converted into electric power is aimed at 25% (a) Assuming 200 MeV of thermal energy to come from each fission event on an average find The number of events that should take place energy day (b) Assuming the fission to take place largely through 235 U at what rate will the amount of 235 U decrease ? Express your answer in kg/day (c) Assuming that uranium enriched to 3% in 235 U will be used how much uranium is needed per month (30 days )? ,d 'kgj dh vkcknh nl yk[k gSA izR;sd O;fDr dh ÅtkZ vko';drk dk vkSlr eku 300 W gSA bl 'kgj dks fo|qr iznku
djus ds fy;s ,d fj,DVj dh jpuk dh tkrh gSA bldh Å"ek ÅtkZ dks fo|qr&ÅtkZ esa ifjofrZr djus dh n{krk dk y{; 25% gS (a) eku yhft;s fd izR;sd fo[k.Mu ls vkSlru 200 MeV Å"ek ÅtkZ mRiUu gksrh gS] izR;sd fnu gksus okys fo[k.Muksa dh la[;k Kkr dhft;sA (b) eku yhft;s fd 235U dh ek=kk fdl nj ls de gksxh\ viuk mÙkj fdxzk@fnu esa nhft;sA (c) eku yhft;s 3% , 235U le`) ;wjfs u;e iz;D q r fd;k tkrk gS] rks izfrekg (30 fnu) fdruk ;wjfs u;e iz;D q r fd;k tk;sxk? Ans:
(a) 3.24 × 10 24 (b) 1.26 kg /day (c) 1263 kg)
50.
Calculate the Q values of the following fusion reactions : fuEufyf[kr fo[k.Mu vfHkfØ;kvksa ds fy;s Q ekuksa dh x.kuk dhft;s (A)
&
2 2 3 1 1 H 1 H 1 H 1H
(B) 12 H 12H 32He n (C) 12 H 13H 24He n Atomic masses are m( 12 H ) = 2.014102 u , m ( 13 H ) = 3.016049 u m( 32 H ) = 3.016029 u, m( 24 H ) = 4.0022603 u. Ans:
(a) 4.05 Me V (b) 3.25 Me V (c) 17.57 Me V
51.
Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 equals the Coulomb potential energy at 2 fm. ghfy;e IykTek esa lay;u ij fopkj dhft;sA og rki Kkr dhft;s ftl ij vkSlr rkih; ÅtkZ 1.5 kT dk eku 2 fm ¼QehZ½
ij dwykeh; fLFkfrt ÅtkZ ds cjkcj gksA 10
Ans :
2.23 × 10
52.
Calculate the Q -value of the fusion reaction 4 He + 4He = 8Be. s such a fusion energetically favorable? Atomic mass of 8Be is 8.0053 u and that of 4He is 4.0026 u. fuEu lay;u vfHkfØ;k ds fy;s Q-eku dh x.kuk dhft;s : 4 He + 4He = 8Be. D;k ;g lay;u ÅtkZ ds vk/kkj ij laHko gS\ 8Be dk ijek.kq Hkkj 8.0053 u vkSj 4He dk 4.0026 u gSA – 93 .1 Ke V , no
Ans: 53.
K
Calculate the energy that can be obtained from 1 kg of water through the fusion reaction 2 H + 2H 3H + P .. Assume that 1.5 × 10 – 2 % of natural water is heavy water D2O (by number of molecules ) and all the deuterium is used for fusion fuEu lay;u vfHkfØ;k }kjk 1 fdxzk ikuh esa mRiUu Å"ek dh x.kuk dhft;s % 2 H + 2H 3H + P .. ;g eku yhft;s fd izkÑfrd ty esa 1.5 × 10 – 2 % Hkkjh ikuh D2O gS (v.kqvksa dh la[;k ds vk/kkj ij) vkSj leLr M~;fw Vfj;e
lay;u esa iz;D q r gks tkrk gSA Ans :
3200 MJ
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