Question

Calculate the wavelength of a photon that is emitted when an electron in the Bohr orbit $n=2$ returns to the orbit $n=1$ in a $H{e}^{+}$ ion.
The ionisation potential of the ground state $H{e}^{+}$ ion is $8.68\times {10}^{-11}\text{erg per atom}$.

• A. $205\stackrel{o}{\text{A}}$
• B. $255\stackrel{o}{\text{A}}$
• C. $305\stackrel{o}{\text{A}}$
• D. $355\stackrel{o}{\text{A}}$

Answer 1

The correct option is C $305\stackrel{o}{\text{A}}$

Ionisation energy, $\Delta E=E_{\infty }-E_1=8.68\times {10}^{-11}\text{erg}$
$\Rightarrow E_1=-8.68\times {10}^{-11}\text{erg}(\because E_{\infty }=0)$
$=-8.68\times {10}^{-18}\text{J}(\because 1\text{J}={10}^7\text{erg})$

For the transition $n=2$ to $n=1$, energy of the emitted radiation, $\Delta E=E_2-E_1$
$=\frac{-8.68\times {10}^{-18}}{n^2}-(-8.68\times {10}^{-18})(\because E_n\propto \frac{1}{n^2})$

$=8.68\times {10}^{-18}[1-\frac{1}{n^2}]$

$=8.68\times {10}^{-18}[1-\frac{1}{4}]=6.51\times {10}^{-18}\text{J}$

$\therefore \lambda =\frac{hc}{\Delta E}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{6.51\times {10}^{-18}}$

$=305\stackrel{o}{\text{A}}$