Question

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

Answer 1

value of R= 109678
wavelength =1/ no of waves
I hope this is the correct answer

Answer 2

Answer: The wavelength of the radiation emitted when an electron in a hydrogen atom makes a transition from an energy level with n=3 to n=2 is 656 nm.

Explanation:

Initial energy level =$n_i=3$

Final energy level =$n_f=2$

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = final energy level

$n_i$=Initial energy level

$\frac{1}{\lambda }=1.0973\times 10^{7} m^{-1}\times [\frac{1}{3^2}-\frac{1}{2^2}]$

$\lambda =-6.56\times 10^{-7}m$

Negative sign means that electron is coming from higher energy level to lower energy level.

$\lambda =6.56\times 10^{-7}m=656 nm$

The wavelength of the radiation emitted when an electron in a hydrogen atom makes a transition from an energy level with n=3 to n=2 is 656 nm.