Question

calculate wave number, wavelength, frequency and energy of radiation emitted by an electron in h spectrum when it jumps from n=4 to n=2​

Answer 1

use rydbergs formula R(1/n1² - 1/n2²) to find the wave number. take reciprocal of wave number we get wave length after finding the wave length use the formula f=c/lamda (lamda is the wave length) and find the frequency now use the equation e=hf (f is the frequency that we found earlier) to find energy. I hope it helps:)

Answer 2

Answer:

As requested by Sudiksha Gupta.

I think you want the numerical data in this answer as you want to know the final answer of this question.

So no problem, I will do it.

Here,

n1 = 1 (Ground Level)

n2 = 3(Excited state)

Rh = Rydberg constant = 10 973 731.6 m-1

λ = wavelength.

v = frequency.

The and formula we will use is :

Putting values :

1/ λ = 10 973 731.6 * (1 - 1/9)

1/ λ = 10 973 731.6 * 8/9

1/ λ = 9754428.09

λ = 1.025 * 10^-7 m

Using v * λ = c (c c= speed of light = 3 * 10^8)

v = c / λ

v = 2.92 * 10^15 sec^-1

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Explanation:

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