Question

The dissociation energy of $H_{2}is430.53kJmo{l}^{-1}.If{H}_2$ is dissociated by illumination with radiation of wavelength $253.7nm$, the fraction of the radiant energy which will be converted into kinetic energy is given by

• A. 8.86%
• B. 2.33%
• C. 1.3%
• D. 90%

Answer 1

The correct option is A

8.86%

From Einstein's Photoelectric effect equation,

KE = $h\nu -h{\nu }_0$ and also,

$1mole=6.022\times {10}^{23}$

Here, $h{\nu }_0$ is dissociation energy.

Therefore $h{\nu }_0=\frac{430.53\times {10}^{23}}{6.023\times {10}^{23}}$

Now,

$KE$ = $h\nu -h{\nu }_0$

$\Rightarrow KE=\frac{430.53\times {10}^3}{6.023\times {10}^{23}}$

$\Rightarrow \frac{hc}{\lambda }=KE=\frac{430.53\times {10}^3}{6.023\times {10}^{23}}$

$\Rightarrow KE=KE=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{253.7\times {10}^{-9}}$ $[\lambda =253.7nm]$

= $\frac{430.53\times {10}^3}{6.022\times {10}^{23}}$

= $6.9\times {10}^{-20}$

$\therefore$ Fraction = $\frac{6.9\times {10}^{-20}}{7.83\times {10}^{-19}}-0.0886$

$\%=0.0886\times 100=8.86\%$