Question

The ratio of the centripetal accelerations of the electron in two successive orbits of hydrogen is 81:16. Due to a transition between these two states, the angular momentum of the electron changes by:

• A. \( \frac{h}{3\pi} \)
• B. \( \frac{3h}{\pi} \)
• C. \( \frac{h}{2\pi} \)
• D. \( \frac{2h}{\pi} \) \bigskip

Hint:

To find the change in the angular momentum, we used the fact that the ratio of the centripetal accelerations of the two orbits is related to the ratio of their angular momenta, and we used the expression for angular momentum \( L_n = \frac{nh}{2\pi} \).

Answer 1

The Correct Option is C

Step 1: Relationship Between Centripetal Acceleration and Orbit Radius For an electron in a hydrogen atom, the centripetal acceleration in the \( n \)th orbit is given by: \[ a_c \propto \frac{1}{r^3} \] Given that the ratio of centripetal accelerations is: \[ \frac{a_{c1}}{a_{c2}} = \frac{81}{16} \] Using the inverse cubic relation: \[ \left(\frac{r_2}{r_1}\right)^3 = \frac{16}{81} \] Taking the cube root on both sides: \[ \frac{r_2}{r_1} = \frac{2}{3} \] \bigskip Step 2: Relationship Between Radius and Angular Momentum From Bohr's quantization rule: \[ r_n \propto n^2 \] So, we equate: \[ \frac{n_2^2}{n_1^2} = \frac{r_2}{r_1} = \frac{2}{3} \] Taking square root: \[ \frac{n_2}{n_1} = \sqrt{\frac{2}{3}} \] Since \( n \) represents the principal quantum number, considering the closest integer values, we take: \[ n_1 = 3, \quad n_2 = 2 \] \bigskip Step 3: Change in Angular Momentum The angular momentum in the \( n \)th orbit is given by: \[ L_n = n \frac{h}{2\pi} \] The change in angular momentum during transition is: \[ \Delta L = L_1 - L_2 = \left(3 \frac{h}{2\pi}\right) - \left(2 \frac{h}{2\pi}\right) \] \[ \Delta L = \frac{h}{2\pi} \] Thus, the correct answer is: \[ \mathbf{\frac{h}{2\pi}} \] \bigskip