Discussion Forum

If the electron in the hydrogen atom jumps from the third orbit to second orbit , the wavelength of the emitted radiation in term  of Rydberg constant

Answer:

Explanation:

 As we know, $\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{\lambda}=R\left(\frac{9-4}{36}\right)=\frac{5R}{36}$

$\therefore$    $\lambda=\frac{36}{5R}$