Practicing Success
If the minimum wavelength corresponding the Paschen series of hydrogen spectra is 820 nm then that corresponding to the Balmer series would be. |
1640 nm 364.4 nm 528.4 nm 298.2 nm |
364.4 nm |
The correct answer is Option (2) → 364.4 nm Wavelength of radiation emitted in hydrogen spectra $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ For Paschen $n_1=3, n_2 \geq 4$ for $\lambda_{\min} n_1=3, n_2=\infty$ $\frac{1}{\lambda_1}=\frac{R}{9}$ For Balmer $n_1=2, n_2 \geq 3$ For $\lambda_{\min } n_1=2, n=\infty$ $\frac{1}{\lambda_2}=\frac{R}{4}$ $\frac{\lambda_2}{\lambda_1}=\frac{4}{9}$ $\lambda_2=\frac{4}{9} \times \lambda_1$ $=\frac{4}{9} \times 820$ $\approx 364.4 ~nm$ |