Here, \(T=30 \ mins.=0.5h, \ t=1.5hr. \ \therefore \ n=\frac { t }{ T } =\frac { 1.5 }{ 0.5 } =3\)
When distance is changed, count rate after 3 half lives is \(5{ s }^{ -1 }.\) Therefore, count rate at the beginning of 1.5 hr. \(={ \left( 2 \right) }^{ 3 }\times 5=40{ s }^{ -1 },\) which is \(\frac { 1 }{ 9 } \times 360{ s }^{ -1 }=\frac { 1 }{ 9 } th\) of original count rate. i.e. Intensity has been reduced by a factor of 9. As intensity \(\propto \frac { 1 }{ { \left( dist \right) }^{ 2 } } ,\)  therefore, distance of source from counter must have increased by a factor of 3 i.e. New distance \(=3\times 2=6m\)

Here, \(\tau =20\quad milli\quad sec=20\times { 10 }^{ -3 }s,C=100\mu F=100\times { 10 }^{ -6 }F.\)
The activity of the sample at time t is \(A={ A }_{ 0 }{ e }^{ -\lambda t }\)
Charge on the capacitor at time t is \(Q={ Q }_{ 0 }{ e }^{ -t/RC }\)
\(\frac { Q }{ A } =\frac { { Q }_{ 0 }{ e }^{ -t/RC } }{ { A }_{ 0 }{ e }^{ -\lambda t } } \)
Clearly, the ratio will be independent of t, if \(\lambda =\frac { 1 }{ RC } \ or \ R=\frac { 1 }{ \lambda C } =\frac { \tau }{ C } =\frac { 20\times { 10 }^{ -3 } }{ 100\times { 10 }^{ -6 } } =200 \ ohm\)

\(As \ \ { N }_{ x }={ N }_{ 0 }{ e }^{ -\left( { E }_{ x }-{ E }_{ 0 } \right) /KT } \ \therefore \ \frac { { N }_{ x } }{ { N }_{ 0 } } ={ e }^{ -\left( { E }_{ x }-{ E }_{ 0 } \right) /KT }\)
\( \ln { \left( \frac { { N }_{ x } }{ { N }_{ 0 } } \right) } =-\frac { \left( { E }_{ x }-{ E }_{ 0 } \right) }{ kT } \ or \ -T=\frac { \left( { E }_{ x }-{ E }_{ 0 } \right) }{ k\ln { \left( \frac { { N }_{ x } }{ { N }_{ 0 } } \right) } }\)
\(Now, \ { E }_{ x }-{ E }_{ 0 }=2.2eV \ and \ \frac { { N }_{ x } }{ { N }_{ 0 } } =1.10; \ k=8.62\times { 10 }^{ -5 }eV/K\)
\(\therefore \ -T=\frac { 2.2eV }{ \left( 8.62\times { 10 }^{ -5 }eV/K \right) \ln { \left( 1.10 \right) } } =2.67\times { 10 }^{ 5 }K \ or \ T=-2.67\times { 10 }^{ 5 }K\)

In an ordinary atom, as a first approximation, we ignore the motion of the nucleus, being too heavy. In a positronium atom, a positron replaces proton of hydrogen atom. As electron and positron masses are equal, the motion of the positron cannot be ignored. We consider motion of electron and positron about their centre of mass. A detailed analysis (beyond the scope of this book) shows that formulae of Bohr model apply to positronium atom provided that we replace \({ m }_{ e }\) by what is known as reduced mass of the electron. For positronium, the reduced mass is \({ m }_{ e }/2.\) In the transition n = 2 to n = 1, the wavelength of radiation emitted is double than that of the corresponding radiation emitted for a similar transition in hydrogen atom, which has a wavelength of \(1217\mathring { A } \) ; and hence is equal to \(2\times 1217=2434\mathring { A } .\) This radiation lies in the ultra-violet part of the electromagnetic spectrum.

The instantaneous activity of a radioactive sample is given by \(A={ A }_{ 0 }{ e }^{ -\lambda t }\)
Differentiating both sides w.r.t. t, we get \(\frac { dA }{ dt } ={ A }_{ 0 }{ e }^{ -\lambda t }\left( -\lambda \right) =-\lambda { A }_{ 0 }{ e }^{ -\lambda t }=-\lambda A=-\lambda \left( \lambda N \right) =-{ \lambda }^{ 2 }N\)
Using \(\lambda =\frac { 0.693 }{ T } , \ we \ get \ \frac { dA }{ dt } =-{ \left( \frac { 0.693 }{ T } \right) }^{ 2 }N, \ i.e., \ \frac { dA }{ dt } \propto \frac { 1 }{ { T }^{ 2 } } ,\) which was to be proved.