Question

The electron in a hydrogen atom jumps back from an excited state to ground state, by emitting a photon of wavelength ${\lambda }_0=\frac{16}{15R}$, 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 such that $\frac{1}{{\lambda }_1}$+$\frac{1}{{\lambda }_2}$+$\frac{1}{{\lambda }_3}=\frac{15R}{16}$
• B. Emitting $2$ photons of wavelengths ${\lambda }_1$,and ${\lambda }_2$ and such that $\frac{1}{{\lambda }_1}$+$\frac{1}{{\lambda }_2}$=$\frac{15R}{16}$
• C. Emitting $2$ photons of wavelengths ${\lambda }_1,{\lambda }_2$ and such that ${\lambda }_1+{\lambda }_2=\frac{15R}{16}$
• D. Emitting $3$ photons of wavelengths ${\lambda }_1,{\lambda }_2$ and such that ${\lambda }_1$+${\lambda }_2$+${\lambda }_3=\frac{15R}{16}$

Answer 1

Answer: (A), (B)

The correct options are
A Emitting $3$ photons of wavelengths ${\lambda }_1,{\lambda }_2$ and such that $\frac{1}{{\lambda }_1}$+$\frac{1}{{\lambda }_2}$+$\frac{1}{{\lambda }_3}=\frac{15R}{16}$
B Emitting $2$ photons of wavelengths ${\lambda }_1$,and ${\lambda }_2$ and such that $\frac{1}{{\lambda }_1}$+$\frac{1}{{\lambda }_2}$=$\frac{15R}{16}$