Question

Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths $\lambda_{N} ,\, \lambda_{A}$ respectively. The ratio $\frac{\lambda_{N}}{\lambda_{A}}$ is closest to :

• A. $10^{-6}$
• B. 10
• C. $10^{-10}$
• D. $10^{-1}$

Answer 1

Answer: (A)

$10^{-6}$

Answer 2

We know that $E=\frac{h c}{\lambda}$
So, for atom $E_{ A }=\frac{h c}{\lambda_{ A }}$
for neutron $E_{ N }=\frac{h c}{\lambda_{ N }}$
Then, $\frac{E_{ A }}{E_{ N }}=\frac{h c}{\lambda_{ A }} \times \frac{\lambda_{ N }}{h c}$
$\Rightarrow \frac{\lambda_{ N }}{\lambda_{ A }}$
Here, $E_{A}$ is order of $eV$ and $E_{ N }$ is order of $MeV$.
Therefore,
$\frac{\lambda_{ N }}{\lambda_{ A }}=\frac{E_{ A }}{E_{ N }}=\frac{ eV }{ MeV }=\frac{1 eV }{10^{6} eV }$
$\lambda_{ N }=10^{-6} \lambda_{ A }$