Question

A photon having wavelength equal to 500 nm is used for photolysis of a molecule X2 having bond energy of 144 kJ/mol. The % of the energy of photon which is converted into the kinetic energy of X atoms is X. The value of is (to the nearest integer) [hc = 1240 eV nm, 1 eV/molecule = 96 kJ/mol] 10 =​

Answer 1

Answer:

The energy of a 500 nm photon is 4 x 10^-19 J.

Explanation:

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Answer 2

Given:

The wavelength of the photon = 500 nm

The bond energy of molecule X₂ = 144 kJ/mol.

To find:

The percentage of the energy of the photon that is converted into kinetic energy of X atoms

Solution:

The incident photon strikes on the molecule X₂. Due to this X₂ dissociates into 2 X atoms by gaining bond energy from the photon.

The rest of the energy of the photon converts into kinetic energy of the atoms.

⇒ Kinetic Energy of the atom = Energy of the incident photon - Bond Energy of X₂

We know that energy of an incident photon = E = hc / λ

Here, hc = 1240 ev nm and λ = 500 nm

Substituting,

E = 1240 / 500 = 2.48 eV/molecule

Since, 1 eV/molecule = 96 kJ/mol

⇒ E = 2.48 × 96 = 238.08 kJ/mol

Bond energy of X₂ = 144 kJ/mol

⇒ Kinetic Energy of X atoms = 238.08 - 144 = 94.08 kJ/mol

Now the percentage energy of the photon that is converted into kinetic energy =  $\frac{KE of X atoms }{Energy of incident photon } X 100$

= 94.08 X 100 / 238.08

= 40% (Approximated)

Hence, 40% of the energy of photons is converted into the kinetic energy of X atoms.