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

Answer:

$4.86\times10^{-7}\ \text{m}$

Explanation:

R = Rydberg constant = $1.09677583\times 10^7\ \text{m}^{-1}$

$n_1$ = Principal quantum number of an energy level = 2

$n_2$ = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

$\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}$

The wavelength of the light emitted is $4.86\times10^{-7}\ \text{m}$.