Question

On absorbing light of 3800 Å, bromine molecule undergoes dissociation and form atoms. The kinetic energy of one quantum of radiation is absorbed by each molecule would be (Bond energy of Br2 =190 kJ/mol)

Answer 1

h is plank’s constant = 6.62 × 10^-34 Js

c is speed of photon i.e. speed of light

Wavelength of light λ = 3800* 10^-10 m

Energy provided to Bromine molecule is given by hc/λ

= 6.62 × 10^-34 × 3 × 10^8 ÷ 3800× 10^-10

= 5.226× 10^-19 J

Energy utilized for breaking of bromine molecule = Bond Energy per mole/ Avagadro number

= 190 × 10^3 ÷ 6.023 × 10 ^23

= 3.15× 10^-19 J

Kinetic energy of bromine molecule= [5.226-3.15] × 10 ^-19 J = 2.067 × 10 ^-19 J

Therefore, Kinetic energy per bromine atom = 2.067 × 10 ^-19 / 2

= 1.034 × 10^-19 J

Answer 2

Answer:

1.04 ×10-¹⁹

Explanation:

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