NEET Physics Question - Modern Physics

Consider a hypothetical hydrogen like atom. Wavelength of spectral lines corresponding to transition of electron from \( n = m \) to \( n = 1 \) are given by \( \lambda = \frac{1800 m^2}{m^2 - 1} \text{ \A^\circ} \) \( (m = 2, 3, 4, 5, \dots, \infty) \). Minimum energy of emitted photon is nearly

5.2 eV

7.03 eV

10.2 eV

13.6 eV

Solution:

Minimum energy corresponds to the maximum wavelength, which occurs at \( m = 2 \). Thus, \( \lambda_{\max} = \frac{1800 \times 2^2}{2^2 - 1} = 2400 \text{ \A^\circ} \). The minimum energy is \( E_{\min} = \frac{12400}{\lambda_{\max}} = \frac{12400}{2400} \approx 5.17 \text{ eV} \approx 5.2 \text{ eV} \).

Rankers Physics

Minimum energy of emitted photon

Solution:

Leave a Reply Cancel reply