CBSE 12th Standard Physics Subject Atoms HOT Questions 3 Mark Questions 2021
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CBSE 12th Standard Physics Subject Atoms HOT Questions 3 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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Show that the instantaneous rate of change of activity of a radioactive substance is inversely proportional to the square of its half life.
(a) -
A positronium atom is a bound state of an electron \(\left( { e }^{ - } \right) \) and its antiparticle, the positron \(\left( { e }^{ + } \right) \) revolving round their centre of mass. In which part of the em spectrum does the system radiate when it de-excites from its first excited state to the ground state?
(a) -
A population inversion for two energy levels is often described by assigning a negative Kelvin temperature to the system. What negative temperature would describe a system in which population of the upper energy level exceeds that of the lower by 10% and the energy difference between the two levels is 2.2 eV?
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A radioactive sample decays with an average life of 20 milli sec. A capacitor of capacitance \(100\mu F\) 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?
(a) -
A point source of \(\gamma \)-radiation has a half life of 30 minutes. The initial count rate, recorded by Geiger counter placed 2 m from the source is \(360{ s }^{ -1 }.\) The distance between the counter and the source is altered. After 1.5 hours, the count rate recorded in \(5{ s }^{ -1 }.\) What is the new distance between the counter and the source?
(a)
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CBSE 12th Standard Physics Subject Atoms HOT Questions 3 Mark Questions 2021 Answer Keys
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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. -
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.
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\(As \ \ { N }_{ x }={ N }_{ 0 }{ e }^{ -\left( { E }_{ x }-{ E }_{ 0 } \right) /KT } \ \therefore \ \frac { { N }_{ x } }{ { N }_{ 0 } } ={ e }^{ -\left( { E }_{ x }-{ E }_{ 0 } \right) /KT }\)
\( \ln { \left( \frac { { N }_{ x } }{ { N }_{ 0 } } \right) } =-\frac { \left( { E }_{ x }-{ E }_{ 0 } \right) }{ kT } \ or \ -T=\frac { \left( { E }_{ x }-{ E }_{ 0 } \right) }{ k\ln { \left( \frac { { N }_{ x } }{ { N }_{ 0 } } \right) } }\)
\(Now, \ { E }_{ x }-{ E }_{ 0 }=2.2eV \ and \ \frac { { N }_{ x } }{ { N }_{ 0 } } =1.10; \ k=8.62\times { 10 }^{ -5 }eV/K\)
\(\therefore \ -T=\frac { 2.2eV }{ \left( 8.62\times { 10 }^{ -5 }eV/K \right) \ln { \left( 1.10 \right) } } =2.67\times { 10 }^{ 5 }K \ or \ T=-2.67\times { 10 }^{ 5 }K\) -
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 \(\lambda =\frac { 1 }{ RC } \ or \ R=\frac { 1 }{ \lambda C } =\frac { \tau }{ C } =\frac { 20\times { 10 }^{ -3 } }{ 100\times { 10 }^{ -6 } } =200 \ ohm\) -
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 \(=3\times 2=6m\)
