Question

The dissociation energy of $ {H_2} $ is $ 430.53{\text{ }}KJ/mol $ . If H2 is exposed to radiant energy of wavelength $ 253.7nm $ , what percentage of radiant energy will be converted into kinetic energy?

Answer 1

Hint: We can use the formula of energy of photon given as-
$ \Rightarrow $ $ E = h \times f $ where frequency f = (speed of light/wavelength) and h is Planck’s constant to find the energy when an incident ray falls on a photon. Now, the hydrogen molecules will start dissociating so we will find the dissociation energy per molecule EH by dividing the given energy by Avogadro number. Now, to find the amount of radiant energy converted into kinetic energy, we will use the formula- $\Rightarrow $ $ K.E. = E - {E_H} $
Now to find the percentage of kinetic energy we will use the formula-
 $ \Rightarrow K.E.\% = \dfrac{{K.E.}}{E} \times 100 $
Put the given values in the formula and solve it to get the answer.

Complete step by step solution:
Given, dissociation energy of $ {H_2} $ = $ 430.53{\text{ }}KJ/mol $
And the wavelength of radiant energy= $ 253.7nm $ = $ 253.7 \times {10^{ - 9}} $ m.
The radiant energy is when an incident on a hydrogen molecule starts dissociating. So we have to find the dissociation energy per molecule, then we will divide the given energy by the Avogadro number which is $ 6.023 \times {10^{23}} $ .
Now dissociation energy per molecule is given as-
 $ \Rightarrow $ EH= $ \dfrac{{430.53 \times {{10}^{ - 3}}}}{{6.023 \times {{10}^{23}}}} $ J = $ 7.15 \times {10^{ - 19}} J {on simplifying}$
Since the wavelength of the radiant energy incident on $ {H_2} $ is $ 253.7 \times {10^{ - 9}} $ and we know that planck's constant= $ 6.626 \times {10^{ - 34}} $ and speed of light= $ 3.0 \times {10^8} $
Then we can write the energy of photon as-
$ \Rightarrow $ $ E = h \times f$
On putting the given values, we get-
$ \Rightarrow $ E= $ \dfrac{{6.626 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{253.7 \times {{10}^{ - 9}}}} = 7.83 \times {10^{ - 19}} J {on simplifying}$
Now, Energy converted into kinetic energy = Energy left after dissociation of the bond
$\Rightarrow $ $ K.E. = E - {E_H}$
On putting values, we get-
$ \Rightarrow $ K.E. = $ \left( {7.83 - 7.15} \right) \times {10^{ - 19}} $ = $ 0.68 \times {10^{ - 19}} $
Now that we know the value of K.E., we can find its percentage.
The percentage of energy converted into K.E. is given as-
$\Rightarrow K.E.\% = \dfrac{{K.E.}}{E} \times 100$
On putting the given values, we get-
$ \Rightarrow K.E.\% $ = $ \dfrac{{0.68 \times {{10}^{ - 19}}}}{{7.83 \times {{10}^{ - 19}}}} \times 100 = 8.68 %$
The percentage of energy converted into K.E. is $ 8.68\% $

Note:
Here, the student may get confused as to why we are subtracting dissociation energy from the energy of the photon. When radiant energy is incident on the molecule it starts dissociating due to which the molecule possesses kinetic energy. This is due to the movements of molecules after dissociation. We calculate this kinetic energy by subtracting the energy of dissociation (per molecule) from the photon’s energy.