Suppose potential energy between electron and proton at separation $$r$$ is given by $$U = K\ln \left( r \right),$$ where $$K$$ is a constant. For such a hypothetical hydrogen atom, the ratio of energy difference between energy levels ($$n = 1$$ and $$n= 2$$ ) and ($$n=2$$ and $$n=4$$ ) is
A.
1
B.
2
C.
3
D.
4
Answer :
1
Solution :
$$ - \frac{{dU}}{{dr}} = F\,\,\left( {{\text{conservative force field}}} \right)$$
$$ \Rightarrow F = \frac{{ - K}}{r}$$ provides the centrifugal force for circular motion of electron.
$$\frac{{m{v^2}}}{r} = \frac{K}{r} \Rightarrow r = \frac{{nh}}{{2\pi \sqrt {mK} }}$$
$$K.E.$$ of electron $$ = \frac{1}{2}m{v^2} = \frac{1}{2}K$$
$$P.E.$$ of electron $$ = K\ln r$$
$$\eqalign{
& E\left( n \right) = {\text{Total}}\,{\text{energy}} = K.E. + P.E. \cr
& = \frac{1}{2}K + K\ln \,r = \frac{K}{2}\left[ {1 + \log \frac{{{n^2}{h^2}}}{{4{\pi ^3}mk}}} \right] \cr} $$
Required ratio $$ = \frac{{E\left( 2 \right) - E\left( 1 \right)}}{{E\left( 4 \right) - E\left( 2 \right)}} = 1$$
Releted MCQ Question on Modern Physics >> Atoms And Nuclei
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