Question

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 $\rightarrow$ n = 3 and n = 4 $\rightarrow$ n = 6 transitions, respectively, is:

• A. 1 : 9
• B. 1 : 36
• C. 1 : 4
• D. 1 : 25

Hint:

For hydrogen atom transitions, the wavelength is inversely proportional to the energy difference, which can be calculated using the Rydberg formula. The energy difference determines the wavelength of light absorbed or emitted.

Answer 1

The Correct Option is C

The wavelength (\(\lambda\)) of light absorbed or emitted during a transition in a hydrogen atom can be related to the energy difference between the two states by the equation: \[ \Delta E = \frac{R_H}{n_1^2} - \frac{R_H}{n_2^2} \] where \( R_H \) is the Rydberg constant, \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final orbits, respectively. The energy and wavelength are inversely related: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck's constant and \( c \) is the speed of light.
Thus, the ratio of the wavelengths is the inverse of the ratio of the energies.
Therefore, we calculate the energy differences for the two transitions: 1. For the n = 2 \(\rightarrow\) n = 3 transition: \[ \Delta E_1 = \frac{R_H}{2^2} - \frac{R_H}{3^2} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \times \frac{5}{36} \] 2. For the n = 4 \(\rightarrow\) n = 6 transition: \[ \Delta E_2 = \frac{R_H}{4^2} - \frac{R_H}{6^2} = R_H \left( \frac{1}{16} - \frac{1}{36} \right) = R_H \times \frac{5}{144} \] Now, the ratio of the wavelengths (\( \lambda_1 / \lambda_2 \)) is the inverse of the ratio of the energy differences: \[ \frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{\frac{5}{144}}{\frac{5}{36}} = \frac{36}{144} = \frac{1}{4} \]
Thus, the ratio of the wavelengths is 1 : 16.
Therefore, the correct answer is (3) 1 : 4.