Question

the dissociation energy of H2 is 150kJmol^-1.If H2 is exposed to light energy of wavelength 500 nm,what % of light energy will be converted into kinetic energy?​

Answer 1

Answer:

87.15 %.

Explanation:

We know that the bond dissociation energy, which is known as the H-H bond energy which is required for breaking the H2 molecule into hydrogen atoms.

We know the formulae of radiant energy which is :-

hc/λ=6.626×10^−34×3×10^8/124×10^−9 m= 1.6×10^−18.

Hence, radiant energy per mol will be 1.6×10^−18 x 6.022 x 10^23 mol-1 =965365.4 J mol-1

.

In kilo joule we will get radiant energy per mol = 965.36 KJ mol-1.

Again, we know that K.E. is the Radiant energy - Dissociation energy

.

     =  965.36 - 124

.

     = 841.36 KJ mol-1

.

So, the percentage of the radiant energy converted will be  K.E. is = 841.36/965.36×100 which on calculating we will get the value of radiant energy as 87.15 %.

Answer 2

Answer:

CORRECT ANSWER = 37.5%

First find dissociation energy of H₂ for 1 atom

Dissociation energy of H₂ for 1 atom = 150 × 10³ / Nₐ

Nₐ (Avogadro number)

Dissociation energy of H₂ for 1 atom = 24.9 × 10⁻²⁰ J

Applying E = hc/λ  

We get,

E = 4 × 10⁻¹⁹

∴ Light energy converted = (4 × 10⁻¹⁹) - (24.9 × 10⁻²⁰ J)

                                            = 1.5 × 10⁻¹⁹

∴ % converted =  (4 × 10⁻¹⁹)/1.5 × 10⁻¹⁹ × 100 = 37.5%