Question:medium

In the atomic spectrum of hydrogen, the wavelengths of the spectral lines corresponding to electronic transitions (i) n = 4 to n = 2 and (ii) n = 3 to n = 1 are $\lambda_1$ and $\lambda_2$ \AA{}respectively. The value of $(\lambda_1 - \lambda_2)$ (in cm) is ($R_H$ = Rydberg constant)

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The Rydberg formula, $\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$, is fundamental for calculating wavelengths in atomic spectra. For hydrogen, Z=1. Remember to find $\lambda$ by taking the reciprocal of the result.
Updated On: Mar 30, 2026
  • $\frac{1}{R_H}\left[\frac{24}{101}\right]$
  • $R_H\left[\frac{24}{101}\right]$
  • $\frac{1}{R_H}\left[\frac{101}{24}\right]$
  • $R_H\left[\frac{101}{24}\right]$
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The Correct Option is C

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