Question

Calculate the wavelength of radiation emitted when an electron in a hydrogen atom makes a transition from an energy level with n= 3 to a level with n= 2:

• A. 756 nm
• B. 656 nm
• C. 556 nm
• D. 663 nm

Answer 1

The correct option is B 656 nm

$\frac{1}{\lambda }=109678(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\frac{1}{\lambda }=109678(\frac{1}{2^2}-\frac{1}{3^2})$
$\frac{1}{\lambda }=109678\left(0.139\right)=15233c{m}^{-1}$
$\lambda =6.56\times {10}^{-5}cm=656nm$
Hence, the wavelength of the emitted radiation is 656nm.