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$\dfrac{{{\lambda _N}}}{{{\lambda _A}}}$ is closest to
A 10$^{ - 6}$
B 10
C 10$^{ - 1}$
D 10$^{ - 10}$

Answer 1

When the energy increases the wavelength decreases and vice versa. E = h · c / λ. The nucleus emits radiation of the order 0.1A$^\circ$. The energy of Gamma radiation is in order of MeV. Atoms emit radiation of order ${10^5}$ A$^\circ$. The energy of a hydrogen like atom is in order of eV.

Complete step by step answer:
We know that the energy can be written in the form of wavelength as E=$\dfrac{{hc}}{\lambda }$. When energy increases the wavelength decreases and vice versa.
Therefore the ratio of wavelength can be written as the ratio of energy.
$\dfrac{{{\lambda _N}}}{{{\lambda _A}}} = \dfrac{{{E_A}}}{{{E_N}}}$
The energy of the nuclear radiation is in the order of Mev and that of hydrogen like atom is in the order of eV. The radiation from the nucleus is of order 0.1 A$^\circ$ and that of the atom is of the order 10$^5$ A$^\circ$.
Therefore the required ratio becomes $\dfrac{{{\lambda _N}}}{{{\lambda _A}}} = \dfrac{{0.1}}{{{{10}^5}}} = {10^{ - 6}}$
Or, $\dfrac{{{\lambda _N}}}{{{\lambda _A}}} = \dfrac{{{E_A}}}{{{E_N}}}$
=$\dfrac{{1eV}}{{1MeV}}$ [1MeV=10$^6$eV]
=10$^{ - 6}$

So, the correct answer is “Option A”.

Additional Information:
When a nucleus de-excites energy is emitted in the form of Gamma radiation. The wavelength is in the order of 0.1 A$^\circ$ and the energy associated with it is in the order of 1MeV. When hydrogen-like atoms de-excites wavelength of order 10$^5$ A$^\circ$ and the energy associated with it is in the order of MeV.

Note:
Since the energy goes up as the frequency increases, the energy is directly proportional to the frequency. Because frequency and wavelength are related by a constant (c) the energy can also be written in terms of wavelength: E = h · c / λ.