Question

What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answer 1

Answer:

$\lambda=1282nm$

Explanation:

The wavelength of the photons emitted due to an  atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

$\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2}})$

Here $R_H$ is the Rydberg constant for hydrogen and $n_1,n_2$ are the lower and higher quantum number for the energy levels of the  atomic electron transition, respectively. Replacing the given values and solving for $\lambda$

$\frac{1}{\lambda}=1.097*10^7m^{-1}(\frac{1}{3^2}-\frac{1}{5^2}})\\\frac{1}{\lambda}=7.81*10^5m^{-1}\\\lambda=\frac{1}{7.81*10^5m^{-1}}\\\lambda=1.282*10^{-6}m\\\lambda=1.282*10^{-6}m*\frac{1nm}{10^{-9}m}\\\lambda=1282nm$