$(a)\;2000A^{0}\qquad(b)\;3000A^{0}\qquad(c)\;4000A^{0}\qquad(d)\;5000A^{0}$

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Answer : (a) $\;2000A^{0}$

Explanation :

$\varepsilon=\large\frac{hc}{\lambda}$

For given condition $\;\lambda=\large\frac{1500P^{2} }{P^{2}-1}$

Hence $\;\varepsilon=\large\frac{hc}{\lambda}=\large\frac{hc}{1500}\;(1-\large\frac{1}{P^2})\times10^{10} J$

$=\large\frac{hc}{(1500)\;(1.6\times10^{-19})}\;(1-\large\frac{1}{P^2}) \times10^{10}eV$

$=8.28\;(1-\large\frac{1}{P^2})eV$

Hence energy of $\;n^{th}\;$ state is $\;\varepsilon_{n}=\large\frac{8.28}{n^2}eV$

Maximum energy is released for transition from $\;P=\infty\;$ to $\;P=1\;$ ; hence wavelength of most energetic photon is 1500 $A^{0}\;.$

Least energy is released for transition from $\;n=2\; to \;n=1\;$ transition for P=2 $\;\lambda=2000A^{0}\;.$

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