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. ${10}^{-1}$
• D. ${10}^{-10}$

Answer 1

Answer: (A)

${10}^{-6}$

$\frac{E_A}{E_B}=\frac{\frac{hc}{{\lambda }_A}}{\frac{hc}{{\lambda }_N}}$

So, $\frac{E_A}{E_N}=\frac{{\lambda }_N}{{\lambda }_A}$

As order of $E_A=ev$

and order of $E_N=Mev$

$\therefore \frac{{\lambda }_N}{{\lambda }_A}=\frac{1ev}{1Mev}=\frac{1}{{10}^6}={10}^{-6}$

So option A is correct Answer