Question

The electron in a hydrogen atom jumps back from an excited state to ground state, by emitting a photon of wavelength $\lambda_{0}=\frac{16}{15 R }$, where $R$ is Rydberg's constant. In place of emitting one photon, the electron could come back to ground state by

• A. Emitting $3$ photons of wavelengths $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$ such that $\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}+\frac{1}{\lambda_{3}}=\frac{15 R}{16}$
• B. Emitting $2$ photons of wavelength $\lambda_{1}$ and $\lambda_{2}$ such that $\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}=\frac{15 R }{16}$
• C. Emitting $2$ photons of wavelength $\lambda_{1}$ and $\lambda_{2}$ such that $\lambda_{1}+\lambda_{2}=\frac{16}{15 R }$
• D. Emitting $3$ photons of wavelength $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$ such that $\lambda_{1}+\lambda_{2}+\lambda_{3}=\frac{16}{15 R}$

Answer 1

Answer: (A), (B)

Using the Rydberg's Formula for the given transition
$\frac{1}{\lambda_{0}}=R\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right]$
solving we get $n=4$. Thus electron can come to ground state by emitting two or three photons such that the sum of energies of each transition remain same so we can use
$\frac{h c}{\lambda_{0}}=\frac{h c}{\lambda_{1}}+\frac{h c}{\lambda_{2}}$
$\Rightarrow \frac{1}{\lambda_{0}}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}$
and $\frac{h c}{\lambda_{0}}=\frac{h c}{\lambda_{1}}+\frac{h c}{\lambda_{2}}+\frac{h c}{\lambda_{3}}$
$\Rightarrow \frac{1}{\lambda_{0}}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}+\frac{1}{\lambda_{3}}$