Question

The following diagram indicates the energy levels of $a$ certain atom when the system moves from $4E$ level to $E$ . A photo of wavelength $\lambda _{1}$ is emitted. The wavelength of photon produced during it's transition from $\frac{7}{3}E$ level to $E$ is $\lambda _{2}.$ The ratio $\frac{\lambda _{1}}{\lambda _{2}}$ will be

• A. $\frac{9}{4}$
• B. $\frac{4}{9}$
• C. $\frac{3}{2}$
• D. $\frac{7}{3}$

Answer 1

Answer: (B)

Transition from $4E$ to $E\left(\right.4E-E\left.\right)=\frac{hc}{\left(\lambda \right)_{1}}\Rightarrow \left(\lambda \right)_{1}=\frac{hc}{3 E}\ldots \ldots \ldots \left(\right.i\left.\right)$
Transition from $\frac{7}{3}E$ to $E\left(\frac{7}{3} E - E\right)=\frac{hc}{\left(\lambda \right)_{2}}\Rightarrow \left(\lambda \right)_{2}=\frac{3 hc}{4 E}\ldots .\left(\right.ii\left.\right)$
From equation $\left(\right.i\left.\right)$ and $\left(\right.ii\left.\right)$ $\frac{\lambda _{1}}{\lambda _{2}}=\frac{4}{9}$