CBSE 12th Standard Physics Subject Nuclei HOT Questions 3 Mark Questions With Solution 2021
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CBSE 12th Standard Physics Subject Nuclei HOT Questions 3 Mark Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Physics
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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) -
Two radioactive nuclei A and B, in a given sample disintegrates into a stable nucleus C. At time t = 0, number of A species are 4N0 and that of B are N0. Half-life of A (for conversion to C) is 1 min whereas, that of B is 2 min. Initially, there are no nuclei of C present in the sample. When number of nuclei of A and B are equal, then what would be the number of nuclei of C present in the sample?
(a) -
(a) Deduce the expression ,N = \({ N }_{ 0 }{ e }^{ -\lambda t }\), for the law of radioactive decay
(b) (i) Write symbolically the process expressing the \({ \beta }^{ + }\) decay of \(_{ 11 }^{ 22 }{ Na }\) . Also write the basic nuclear process underlying this decay.
(ii) Is the nucleus formed in the decay of the nucleus \(_{ 11 }^{ 22 }{ Na }\), an isotope or an isobar?(a) -
Write symbolically the process expressing the \({ \beta }^{ + }\) decay of \(_{ 11 }^{ 22 }{ Na }\). Also write the basic nuclear process underlying this decay.
(a) -
Is the nucleus formed in the decay of the nucleus \(_{ 11 }^{ 22 }{ Na }\), an isotope or an isobar?
(a)
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CBSE 12th Standard Physics Subject Nuclei HOT Questions 3 Mark Questions With Solution 2021 Answer Keys
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\({ N }_{ 1 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 1 }t },\quad { N }_{ 2 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 2 }t }\)
\(\\ \frac { { N }_{ 1 } }{ { N }_{ 2 } } ={ e }^{ \left( { \lambda }_{ 2 }-{ \lambda }_{ 1 } \right) t }\quad or\quad t=\frac { \log _{ e }{ \left( { N }_{ 1 }{ /N }_{ 2 } \right) } }{ { \lambda }_{ 2 }-{ \lambda }_{ 1 } } =\frac { \log _{ e }{ 140/1 } }{ \log _{ 2 }{ 2\left[ \frac { 1 }{ { T }_{ 2 } } -\frac { 1 }{ { T }_{ 1 } } \right] } } =\frac { \log _{ e }{ 140 } }{ \log _{ 10 }{ 2 } } \left( \frac { { T }_{ 2 }{ T }_{ 1 } }{ { T }_{ 1 }-{ T }_{ 2 } } \right) \)
On putting the values, we get \(t=6\times { 10 }^{ 9 }\ years\) -
Give, at t = 0, number of nuclei of A = 4N0 and number of nuclei of B = N0.
Half-life of A, TA = 1 min, half-life of B, TB = 2 min
After time t, number of nuclei of A,
\({ n }_{ A }=4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/{ T }_{ A } }\)
\(\Rightarrow \) \({ n }_{ A }=4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/1 }\)
After time t, number of nuclei of B,
\({ n }_{ B }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/{ T }_{ A } } = { n }_{ B }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/2 }\)
Let the number of nuclei of A and B in given sample be after time t, then nA = nB
or \(4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/1 }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/2 }\) or \({ \left( \frac { 1 }{ 2 } \right) }^{ t/2 }={ \left( \frac { 1 }{ 4 } \right) }={ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)
or \(\frac { t }{ 2 } =2\quad \Rightarrow \quad t=4min\)
\(\therefore \ \ { n }_{ A }=4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 4/1 }=\frac { { N }_{ 0 } }{ 4 } \)
and \({ n }_{ B }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 4/2 }=\frac { { N }_{ 0 } }{ 4 } \)
\(\therefore \) Population of C in the sample is
\(=\left( 4{ N }_{ 0 }-\frac { { N }_{ 0 } }{ 4 } \right) +\left( \quad { N }_{ 0 }-\frac { { N }_{ 0 } }{ 4 } \right) =\frac { { 9N }_{ 0 } }{ 2 } \) -
(a) \(\frac { dN }{ dt } =-\lambda N\)
\(\int _{ { N }_{ 0 } }^{ N }{ \frac { dN }{ N } =\int _{ 0 }^{ t }{ -\lambda dt } } \)
\(\left[ { log }_{ e }^{ N } \right] _{ { N }_{ 0 } }^{ N }=-\lambda \left[ t \right] _{ 0 }^{ t }\)
\(loge\frac { N }{ { N }_{ 0 } } =-\lambda t\)
\(N={ N }_{ 0 }{ e }^{ -\lambda t }\)
(b) (i) \(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Ne+{ e }^{ x }+\upsilon \)
Also accept,if a student does not identify the product nucleus and writes as
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Xe+{ e }^{ x }+\upsilon \)
Basic process
\(p\rightarrow n+{ e }^{ + }+\upsilon \)
(ii) Isobar -
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Ne+{ e }^{ x }+\upsilon \)
Also accept, if a student does not identify the product nucleus and writes as
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Xe+{ e }^{ x }+\upsilon \)
Basic process
\(p\rightarrow n+{ e }^{ + }+\upsilon \) -
Isobar
