Question

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Answer 1

According to Formula :
Wave number=ν=R[1/n₁²  -1/n₂²]
where 
R=109678 cm⁻¹n1=2n2=4
ν=109678[1/2² - 1/4²]
  =109678[(4-1)/16]
   =109678x3/16
   
As we know that wave number=1/Wavelength
⇒ ν=1/λ
    λ=1/ν
      =1/[109678x3/16]
       =16/109678x3
       =486x10⁻⁷ cm
       =486x 10⁻⁹m
      = 486 nm

∴wavelength of light emitted is 486 nm

Answer 2

Given:

$n_1=2,n_2=4$

To find:

Wavelength of light emitted

Solution:

We know that

Wave number=$\frac{1}{\lambda}=R(\frac{1}{n^2_1}-\frac{1}{n^2_2})$

Where R=Rydberg constant=$1.097\times 10^{7}/m$

Using the formula

$\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{4}-\frac{1}{16})$

$\frac{1}{\lambda}=1.097\times 10^7(\frac{4-1}{16}$

$\frac{1}{\lambda}=1.097\times 10^7\times \frac{3}{16}$

$\frac{1}{\lambda}=0.2058\times 10^{7}/m$

$\lambda=\frac{1}{0.2058\times 10^{7}}m$

Wavelength,$\lambda=486\times 10^{-9} m=486 nm$

$1nm=10^{-9} m$