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Q. On absorbing light of wavelength 3800 Å , bromine molecule undergoes dissociation and forms 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 = 190 KJ/mol)

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Answer 1

Planck's Quantum Hypothesis states that Light is made of packets of energy called Quanta [singular: Quantum]

So, we are saying something like light is made of particles.

Let us see the incident light first. We are given a light of wavelength $3800 \, \, \AA$

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

$E=hf$

But we are given wavelength, not frequency, so we use another well-known relation:

$\displaystyle c=\lambda f \\ \\ \\ \implies f = \frac{c}{\lambda}$

where

$c$ = Speed of light in vacuum

$\lambda$ = Wavelength

$f$ = Frequency

So we have:

$\displaystyle E=hf \\ \\ \\ \implies \boxed{E=\frac{hc}{\lambda}}$

This gives us the Energy of one quantum of radiation of wavelength $\lambda$

Our data is:

$h$ = Planck's Constant = $6.626 \times 10^{-34} \, \, Js$

$c$ = $3\times 10^8 \, \, m/s$

$\lambda$ = $3800 \, \, \AA = 3800 \times 10^{-10} \, \, m$

So,

$\displaystyle E = \frac{hc}{\lambda} \\ \\ \\ \implies E = \frac{6.626\times 10^{-34} \times 3 \times 10^8}{3800 \times 10^{-10}} \\ \\ \\ \implies \boxed{E \approx 5.231 \times 10^{-19} \, \, J}$

So, we know that an energy of $5.231 \times 10^{-19} \, \, J$ is possessed by one quantum of incident light.

____________________

Now, we are also given the Bond Energy of $Br_2$.

Bond Energy means the energy required to break bonds of one mole of molecules.

It is given as 190 kJ/mol, which means that 190 kJ energy is required to break the bonds of 1 mole of $Br_2$ molecules.

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

$\text{1 mole moecules \rightarrow 190 kJ} \\ \\ \\ \therefore 6.022\times 10^{23} \text{ molecules \rightarrow 190 \times 10^3 J} \\ \\ \\ \therefore \text{1 molecule }\rightarrow \frac{190\times 10^3}{6.022 \times 10^{23}} \\ \\ \\ \implies \boxed{\text{Bond Energy of one molecule } \approx 3.155 \times 10^{-19} \, \, J}$

Now, when a quantum of light is absorbed by a molecule, the bond energy of one molecule is absorbed, which breaks the bond. However, the quantum of light also has some extra energy.

This extra energy is used up in providing Kinetic Energy to the dissociated molecules.

[This is similar to the Photoelectric Effect]

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

Energy of One Photon = $5.231 \times 10^{-19} \, \, J$

Energy required to break bond of one molecule = $3.155 \times 10^{-19} \, \, J$

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

$E' = (5.231\times 10^{-19})-(3.155 \times 10^{-19}) \, \, J \\ \\ \\ \implies E' = 2.076 \times 10^{-19} \, \, J$

This energy left becomes the Kinetic Energy of the dissociated atoms.

Note that one $Br_2$ molecule gives 2 Br atoms. So, this remaining energy is divided equally among both atoms.

$Br_2 \rightarrow 2 \, Br$

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

$\displaystyle K = \frac{E'}{2} \\ \\ \\ \implies K = \frac{2.076 \times 10^{-19} \, \, J}{2} \\ \\ \\ \implies \boxed{\boxed{\bold{K=1.038 \times 10^{-19} \, \, J}}}$

Thus, The Kinetic Energy possessed by one Bromine atom is approximately $\bold{1.038 \times 10^{-19} \, \, J}$

Finally, Let us Summarize what we did:

-> We found the energy of one quantum of light

-> We calculated the energy required to break the bond of one molecule of Bromine

-> When one quantum of light is absorbed by a molecule, the energy required to break the bond is used up, and the molecule dissociates.

-> The remaining energy is what provides the free atoms their Kinetic Energy.

Hope it helps :)

Answer 2

Answer:

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