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

The Correct Option is A

To solve this problem, we need to understand that the nucleus and the atom transition from their first excited states to ground states and emit photons of wavelengths \( \lambda_{N} \) and \( \lambda_{A} \) respectively.

The energy of the photon emitted during the transition can be given by the formula:

E = \frac{hc}{\lambda}

where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the emitted photon.

For the nucleus transitioning from its first excited state:

E_N = \frac{hc}{\lambda_N}

For the atom transitioning from its first excited state:

E_A = \frac{hc}{\lambda_A}

In general, the energy levels of a nucleus are far higher than those of an atom because nuclear forces are much stronger than electromagnetic forces. This implies:

E_N \gg E_A

Thus, we can say:

\frac{1}{\lambda_N} \approx 10^{6} \times \frac{1}{\lambda_A}

Simplifying the above expression for the ratio of wavelengths, we get:

\frac{\lambda_N}{\lambda_A} \approx 10^{-6}

Therefore, the ratio \(\frac{\lambda_{N}}{\lambda_{A}}\) is closest to 10^{-6}.

This matches with option 10^{-6}, confirming this as the correct answer.