Question

Q 14/30 In which of the following transition, emitted
frequency of photon is minimum in Hatom?
O 4th orbit to 3rd orbit
5th orbit to 4th orbit
3rd orbit to 2nd orbit
3rd orbit to 1st orbit​

Answer 1

Answer:

5th orbit to 4th orbit.

thanks

Answer 2

5th orbit to 4th orbit emitted  frequency of photon is minimum in H atom

Therefore, option (2) is correct.

Explanation:

The frequency of emission is

$\nu={\Delta E}{h}$

Where,

$\Delta E=R(\frac{1}{n_1^2}-\frac{1}{n_2^2}})$

Therefore, we can say that when $\Delta E$ is maximum, the frequency will be maximum

(1) When electron jumps from 4th orbit to 3rd orbit

$n_1=3, n_2=4$

$\Delta E=R(\frac{1}{3^2}-\frac{1}{4^2}})$

$\implies \Delta E=R(\frac{1}{9}-\frac{1}{16}})$

$\implies \Delta E=R\frac{7}{144}$

$\implies \Delta E=0.049R$

(2) When electron jumps from 4th orbit to 3rd orbit

$n_1=4, n_2=5$

$\Delta E=R(\frac{1}{4^2}-\frac{1}{5^2}})$

$\implies \Delta E=R(\frac{1}{16}-\frac{1}{25}})$

$\implies \Delta E=R\frac{9}{400}$

$\implies \Delta E=0.023R$

(3) When electron jumps from 3rd orbit to 2nd orbit

$n_1=2, n_2=3$

$\Delta E=R(\frac{1}{2^2}-\frac{1}{3^2}})$

$\implies \Delta E=R(\frac{1}{4}-\frac{1}{9}})$

$\implies \Delta E=R\frac{5}{36}$

$\implies \Delta E=0.139R$

(3) When electron jumps from 3rd orbit to 1st orbit

$n_1=1, n_2=3$

$\Delta E=R(\frac{1}{1^2}-\frac{1}{3^2}})$

$\implies \Delta E=R(\frac{1}{1}-\frac{1}{9}})$

$\implies \Delta E=R\frac{8}{9}$

$\implies \Delta E=0.89R$

Therefore, the lowest frequency of photon is when the electron jumps from 5th orbit to 4th orbit

Hope this answer is helpful.

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