Question

Both the Nucleus and the Atom of some elements are in their first excited states. They get de-excited by emitting photons of wavelengths${\mathrm{\lambda }}_{\mathrm{N}}$, ${\mathrm{\lambda }}_{\mathrm{A}}$ respectively. The wavelengths ${\lambda }_N$$/$${\mathrm{\lambda }}_{\mathrm{A}}$ are closest to:

Answer 1

Step 1: Finding the relation between Energy and wavelength of the emitted photons:

• the wavelength emitted by the Nucleus$=$${\mathrm{\lambda }}_{\mathrm{N}}$
• the wavelength emitted by Atom$=$${\mathrm{\lambda }}_{\mathrm{A}}$

We know, Energy[$\mathrm{E}$] $=$$\frac{hc}{\lambda }$

• So, Energy emitted by the nucleus$=$${\mathrm{E}}_{\mathrm{N}}$
• The energy emitted by the atom$=$${\mathrm{E}}_{\mathrm{A}}$

Where, $E_{\mathrm{N}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{N}}}}$

and, $E_{A=\frac{hc}{{\lambda }_A}}$

Step 2: Relation between the ratios of Energy and Wavelength :

• Then, $\frac{E_A}{E_N}$$=$$\frac{hc}{{\lambda }_A}\times \frac{{\lambda }_N}{hc}$
• $\frac{E_A}{E_N}=\frac{{\lambda }_N}{{\lambda }_A}$

Here, $E_A$ [Energy emitted by the atom ] is in order$\mathrm{eV}$, and ${\mathrm{E}}_{\mathrm{N}}$[ Energy emitted by the nucleus] is in order$\mathrm{MeV}$, where{$M$[Mega]$={10}^6$}.

Step 3: Finding the value of the Ratio :

• $\frac{{\lambda }_N}{{\lambda }_A}=\frac{E_A}{E_N}=\frac{eV}{MeV}=\frac{1}{{10}^6}$
• $\frac{{\mathrm{\lambda }}_{\mathrm{N}}}{{\mathrm{\lambda }}_{\mathrm{A}}}={10}^{-6}$