CBSE 12th Standard Physics Subject Nuclei HOT Questions 3 Mark Questions 2021
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CBSE 12th Standard Physics Subject Nuclei HOT Questions 3 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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(a) If \(\alpha \)-decay of \(_{ 92 }{ { U }^{ 238 } }\) is energetically allowed (i.e. the decay products have a total mass less than the mass of \(_{ 92 }{ { U }^{ 238 } }\)), what prevents \(_{ 92 }{ { U }^{ 238 } }\) from decaying all at once? Why is its half life so large?
(b) The \(\alpha \)-particle faces a Coulomb barrier. A neutron being uncharged faces no such barrier. Why does the nucleus \(_{ 92 }{ { U }^{ 238 } }\) not decay spontaneously, by emitting a neutron?(a) -
The half-lives of radioactive nuclides that emit \(\alpha \)-rays vary from microsecond to billion years. What is the reason for this large variation in the half life of alpha emitters?
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There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutrons is 700 seconds, what fraction of neutrons will decay before they travel a distance of 10 m? Given mass of neutron \(=1.675\times { 10 }^{ -27 }kg.\)
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The radioactive decay rate of a radio active element is found to be \({ 10 }^{ 3 }\) disintegrations/sec. at a certain time. If half life of the element is one second, what would be the decay rate after 1 sec. and after 3 sec.?
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The isotopes of \({ U }^{ 238 }\ and\ { U }^{ 235 }\) occur in nature in the ratio 140 : 1. Assuming that at the time of earth's formation, they were present in equal ratio, make an estimate of the age of the earth. The half lives of \({ U }^{ 238 }\ and\ { U }^{ 235 }\) are \(4.5\times { 10 }^{ 9 }\) years and \(7.13\times { 10 }^{ 8 }\) years respectively.
Given : \(\log _{ 10 }{ 140 } =2.1461,\log _{ 10 }{ 2 } =0.3010\)(a)
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CBSE 12th Standard Physics Subject Nuclei HOT Questions 3 Mark Questions 2021 Answer Keys
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(a) As explained in theory, \(\alpha \)-decay is caused by the quantum mechanical tunnelling of an alpha particle through a repulsive Coulomb barrier. The rate of tunnelling would depend upon the height and width of the barrier. The decay cannot be all at once. And that is the reason why half life of \(_{ 92 }{ { U }^{ 238 } }\) against \(\alpha \)-decay is large.
(b) The possible nuclear reaction is \(_{ 92 }{ { U }^{ 238 } }\rightarrow _{ 92 }{ { U }^{ 237 } }+_{ 0 }{ { n }^{ 1 } }\)
The data shows that \(m\left( _{ 92 }{ { U }^{ 237 } } \right) +m\left( n \right) \) is greater than \(m\left( _{ 92 }{ { U }^{ 238 } } \right) .\) Therefore, the decay is not allowed energetically. Rather, some external energy has to be supplied to separate a neutron from \(_{ 92 }{ { U }^{ 238 } }\). -
The radioactive decay is governed by the law of mass action. There is absolutely no way to predict whether any given nucleus in a radioactive sample will be among the small number of nuclei that decay during the next moment. All nuclei have the same probability of disintegration. Therefore, half lives of radioactive nuclides that emit \(\alpha \) particles may be in microseconds. However, half lives for alpha decay of most of the alpha unstable nuclei are very long. This has been explained in terms of tunnelling of nucleus.
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\(Here; \ K.E.=\frac { 1 }{ 2 } { m\upsilon }^{ 2 }=0.0327 \ eV=0.0327\times 1.6\times { 10 }^{ -19 }J\)
\(\\ or \ \ \frac { 1 }{ 2 } \times \left( 1.675\times { 10 }^{ -27 } \right) \times { \upsilon }^{ 2 }=0.0327\times 1.6\times { 10 }^{ -19 }\)
\(On \ solving, \ we \ get \ \upsilon \ = \ 2.5\times { 10 }^{ 3 }m/s.\)
\(Time, \ t=\frac { distance }{ velocity } =\frac { 10 }{ 2.5\times { 10 }^{ 3 } } =4\times { 10 }^{ -3 }s\)
\(Now \ N={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/T }or \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ t/T }={ \left( \frac { 1 }{ 2 } \right) }^{ \frac { 4\times { 10 }^{ -3 } }{ 700 } }=0.999952\)
\(\therefore \ Fraction \ of \ neutron \ decayed \ = \ 1-0.999952 \ = \ 0.000048\) -
It is known that radioactive decay rate is directly proportional to the number of nuclei left.
\({ N }_{ 0 }\) corresponds to \({ 10 }^{ 3 }\) (disintegration/sec.)
As half life T = 1 sec., therefore
(i) no. of half lives in 1 sec., n = 1
As \(N={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ n }\quad \therefore \quad N=1000{ \left( \frac { 1 }{ 2 } \right) }^{ 1 }=500\)
\(\therefore \) No. of disintegrations/sec. after one sec.= 500
(ii) no. of half lives in 3 sec. \(=n=\frac { 3 }{ 1 } =3\)
As \(N={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ n }\quad \therefore \quad N=1000{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }=\frac { 1000 }{ 8 } =125\)
\(\therefore \) No. of disintegrations/sec. after three sec. = 125 -
\({ N }_{ 1 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 1 }t },\quad { N }_{ 2 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 2 }t }\)
\(\\ \frac { { N }_{ 1 } }{ { N }_{ 2 } } ={ e }^{ \left( { \lambda }_{ 2 }-{ \lambda }_{ 1 } \right) t }\quad or\quad t=\frac { \log _{ e }{ \left( { N }_{ 1 }{ /N }_{ 2 } \right) } }{ { \lambda }_{ 2 }-{ \lambda }_{ 1 } } =\frac { \log _{ e }{ 140/1 } }{ \log _{ 2 }{ 2\left[ \frac { 1 }{ { T }_{ 2 } } -\frac { 1 }{ { T }_{ 1 } } \right] } } =\frac { \log _{ e }{ 140 } }{ \log _{ 10 }{ 2 } } \left( \frac { { T }_{ 2 }{ T }_{ 1 } }{ { T }_{ 1 }-{ T }_{ 2 } } \right) \)
On putting the values, we get \(t=6\times { 10 }^{ 9 }\ years\)