Question

On absorbing light of wavelength 3800 angstrom bromine molecule undergoes dissociation and form atoms. The kinetic energy of one bromine atom assuming that one quantum of radiation is absorbed by each molecule would be (bond energy of br2 =190kj/mol)

Answer 1

Answer:

The answer si 1.038 x 10^-19 J

Explanation:

The energy possessed by one quantum of light of frequency f is given by:

E=hf

We know that  

f = c/ λ

c = Speed of light in vacuum

lambda = Wavelength

f = Frequency

Thus

1- E = hc/ λ  

h = Planck's Constant = 6.626x10^-34 Js

c = 3 x 10^8 m/s

Lmbda = 3800A = 3800x 10^-10 m

Putting the values in above equation

E = hc/ λ  = 6.626x10^-34 x  3 x 10^8 / 3800x 10^-10

E = 5.231 x 10^-19 J

5.231x  10^-19J is possessed by one quantum of incident light.

Let us find the energy required to break the bond of a single molecule.

1-mole molecule----> 190 KJ

1 Molecule = 190 x 10^3/6.022x 10^23

Bond energy of one molecule = 3.155 x 10^-19J

2- Let us calculate how much energy is left after breaking up of one Br_2 molecule.

Energy of One Photon = 5.231 x 10^-19J

Energy required to break bond of one molecule = 3.155 x 10^-19 J

So, Energy left after breaking bond would be (say it is E' )

E' = (5.231x  10^-19)-(3.155 x 10^-19) J

E' = 2.076 x 10^-19 J

3- Note that one Br-2 molecule gives 2 Br atoms. So, this remaining energy is divided equally among both atoms.

So, if we suppose that The Kinetic Energy possessed by one Br atom is K, then we have:

K = E'/2

  = 2.076 x 10^-19 J/2

k  = 1.038 x 10^-19 J

Thus, The Kinetic Energy possessed by one Bromine atom is approximately 1.038 x 10^-19 J