\({ N }_{ 1 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 1 }t },\quad { N }_{ 2 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 2 }t }\)
\(\\ \frac { { N }_{ 1 } }{ { N }_{ 2 } } ={ e }^{ \left( { \lambda }_{ 2 }-{ \lambda }_{ 1 } \right) t }\quad or\quad t=\frac { \log _{ e }{ \left( { N }_{ 1 }{ /N }_{ 2 } \right) } }{ { \lambda }_{ 2 }-{ \lambda }_{ 1 } } =\frac { \log _{ e }{ 140/1 } }{ \log _{ 2 }{ 2\left[ \frac { 1 }{ { T }_{ 2 } } -\frac { 1 }{ { T }_{ 1 } } \right] } } =\frac { \log _{ e }{ 140 } }{ \log _{ 10 }{ 2 } } \left( \frac { { T }_{ 2 }{ T }_{ 1 } }{ { T }_{ 1 }-{ T }_{ 2 } } \right) \)
On putting the values, we get \(t=6\times { 10 }^{ 9 }\ years\)

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

\(Here; \ K.E.=\frac { 1 }{ 2 } { m\upsilon }^{ 2 }=0.0327 \ eV=0.0327\times 1.6\times { 10 }^{ -19 }J\)
\(\\ or \ \ \frac { 1 }{ 2 } \times \left( 1.675\times { 10 }^{ -27 } \right) \times { \upsilon }^{ 2 }=0.0327\times 1.6\times { 10 }^{ -19 }\)
\(On \ solving, \ we \ get \ \upsilon \ = \ 2.5\times { 10 }^{ 3 }m/s.\)
\(Time, \ t=\frac { distance }{ velocity } =\frac { 10 }{ 2.5\times { 10 }^{ 3 } } =4\times { 10 }^{ -3 }s\)
\(Now \ N={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/T }or \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ t/T }={ \left( \frac { 1 }{ 2 } \right) }^{ \frac { 4\times { 10 }^{ -3 } }{ 700 } }=0.999952\)
\(\therefore \ Fraction \ of \ neutron \ decayed \ = \ 1-0.999952 \ = \ 0.000048\)

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.

(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 \(_{ 92 }{ { U }^{ 238 } }\).