If an electron in a hydrogen atom jumps from the $2^{ ext{nd}}$ orbit to the $1^{ ext{st}}$ orbit,

Question

If an electron in a hydrogen atom jumps from the $2^{	ext{nd}}$ orbit to the $1^{	ext{st}}$ orbit, it emits a photon of wavelength $\lambda$. When an electron jumps from the $3^{	ext{rd}}$ orbit to the $2^{	ext{nd}}$ orbit, the corresponding wavelength of the emitted photon will be

Professor PC Thomas Classes · Answer

(a) : For the $2 	o 1$ transition: $\frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{2^2}ight) = \frac{3R}{4} \implies R = \frac{4}{3\lambda}$. For the $3 	o 2$ transition: $\frac{1}{\lambda'} = R \left(\frac{1}{2^2} - \frac{1}{3^2}ight) = R \left(\frac{1}{4} - \frac{1}{9}ight) = \frac{5R}{36}$. Substituting $R$ into this equation: $\frac{1}{\lambda'} = \frac{5}{36} 	imes \left(\frac{4}{3\lambda}ight) = \frac{20}{108\lambda} = \frac{5}{27\lambda}$. Inverting this gives $\lambda' = \frac{27}{5} \lambda$.