Question

calculate the frequency of light emitted for an electron transition from the sixth to second orbit of the hydrogen atom

Answer 1

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Answer 2

Given

Electron transitions from the sixth to second orbit of the hydrogen atom.

To find

Frequency of light emitted

Solution

$n_1$ = 6

$n_2$ = 2

c = Speed of light = $3\times10^8\ \text{m/s}$

$R_h$ = Rydberg's constant of hydrogen = 10967700 /m

We have the relation

$\dfrac{1}{\lambda}=R_h(\dfrac{1}{n_2^2}-\dfrac{1}{n_1^2})\\\Rightarrow \dfrac{1}{\lambda}=10967700(\dfrac{1}{2^2}-\dfrac{1}{6^2})\\\Rightarrow \dfrac{1}{\lambda}=2437266.67\ \text{m}^{-1}$

Frequency is given by

$f=\dfrac{c}{\lambda}\\\Rightarrow f=3\times10^8\times2437266.67\\\Rightarrow f=7.31\times 10^{14}\ \text{Hz}$

The frequency of the photon is $7.31\times 10^{14}\ \text{Hz}$