Question

The wavelength of photons emitted by electron transition between two similar levels in $H$ and $H{e}^{+}$ are ${\lambda }_1$ and ${\lambda }_2$ resepectively. Then the relation between ${\lambda }_1\text{and}{\lambda }_2$ is:

• A. ${\lambda }_2={\lambda }_1$
• B. ${\lambda }_2=2{\lambda }_1$
• C. ${\lambda }_2={\lambda }_1/2$
• D. ${\lambda }_2={\lambda }_1/4$

Answer 1

The correct option is D ${\lambda }_2={\lambda }_1/4$

According to Planck’s quantum theory, energy emitted by electron due to transition can be given by:
$\Delta E=hv=\frac{hc}{\lambda }...\left(i\right)$
where $h=\text{Planck’s constant}=6.63\times {10}^{-34}\text{J s}$
$v=\text{frequency}$
$\lambda =\text{wavelength}$
$c=\text{velocity of light}$

Given: wavelength of photon in $H$ atom $={\lambda }_1$
wavelength of photon in $H{e}^{+}$ ion $={\lambda }_2$

For hydrogen like species,
Energy of an electron $=-13.6\frac{Z^2}{n^2}eV...\left(ii\right)$

As the transition is between the same set of 2 energy levels, but in different species ($H$ and $H{e}^{+}$ ion), we can conclude from eq. 1 and 2:

$\frac{\Delta E_H}{\Delta E_{H{e}^{+}}}=\frac{Z_H^2}{Z_{H{e}^{+}}^2}=\frac{{\lambda }_2}{{\lambda }_1}$
$(\because Z_H=1;Z_{He}=2)$
${\lambda }_2=\frac{1}{4}{\lambda }_1$