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If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:
A
$\frac{20}{7}\lambda$
B
$\frac{20}{13}\lambda$
C
$\frac{16}{25}\lambda$
D
$\frac{9}{16}\lambda$
Explanation
Ratio of (1/n₁² – 1/n₂²) terms.
Detailed Solution
Transition 3 → 2: wavelength $\lambda$; transition 4 → 3: wavelength $\lambda'$
$\frac{1/\lambda}{1/\lambda'} = \frac{R_HZ^2\left(\frac{1}{2^2} - \frac{1}{3^2}\right)}{R_HZ^2\left(\frac{1}{3^2} - \frac{1}{4^2}\right)} \Rightarrow \frac{\lambda'}{\lambda} = \frac{20}{7}$
$\lambda' = \frac{20\lambda}{7}$
$\frac{1/\lambda}{1/\lambda'} = \frac{R_HZ^2\left(\frac{1}{2^2} - \frac{1}{3^2}\right)}{R_HZ^2\left(\frac{1}{3^2} - \frac{1}{4^2}\right)} \Rightarrow \frac{\lambda'}{\lambda} = \frac{20}{7}$
$\lambda' = \frac{20\lambda}{7}$