Question

An electron in $H$-atom in ground state absorbs $1.5$ times as much energy as the minimum required for its escape (i.e., $13.6eV$) from the atom. Calculate the wavelength of emitted electron.

• A. $\lambda =4.70\times {10}^{-10}m$
• B. $\lambda =61\times {10}^{-9}m$
• C. $\lambda =4.70\times {10}^{-14}m$
• D. $\lambda =2.7\times {10}^{-10}m$

Answer 1

The correct option is A $\lambda =4.70\times {10}^{-10}m$

Energy of $I$ orbit of $H$-atom$=-13.6eV$
Energy absorbed by $H$-atom $=\left(3/2\right)\times 13.6eV$
$\therefore$ Energy used in imparting velocity to electron $=\left(13.6/2\right)eV$
or $\left(\frac{1}{2}\right)m{u}^2=\left(13.6/2\right)\times 1.602\times {10}^{-19}J$
$\therefore u=\sqrt{\frac{13.6\times 1.602\times {10}^{-19}}{9.108\times {10}^{-31}}}=1.547\times {10}^{6}mse{c}^{-1}$
Now $\lambda =\frac{h}{mu}=\frac{6.626\times {10}^{-34}}{9.108\times {10}^{-31}\times 1.547\times {10}^6}$
$\therefore \lambda =4.70\times {10}^{-10}m$