Question

Calculate the shortest wavelength of light capable of dissociating the Cl–F bond in one molecule of chlorine monofluoride if the bond energy, or bond dissociation energy, is 251 kJ/mol .

Answer 1

Answer:

477 nm

Explanation:

Using the expression for the energy as:

$E=\frac {h\times c}{\lambda}$

Where,  

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light

Given, $Energy=251\ kJ/mol=251000\ J/mol$

Also, $N_a=6.023\times 10^{23}\ {mol}^{-1}$

So, Energy for one atom = $\frac{251000}{6.023\times 10^{23}}\ J$

Thus, applying values as:

$\frac{251000}{6.023\times 10^{23}}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$251\times \:10^{26}\lambda=5^{20}\times \:125540965$

$\lambda=4.77\times 10^{-7}\ m=477\times 10^{-7}\ m=477\ nm$

( 1 m = $10^{-9}$ nm )