Question

In Bohr’s model of hydrogen atom, find the percentage change in the radius of its orbit when an electron makes a transition from \( n = 3 \) state to \( n = 2 \) state.

Hint:

The radius of the electron's orbit in the Bohr model is proportional to \( n^2 \), where \( n \) is the principal quantum number. The percentage change in radius between two states is given by the ratio of the difference in radii to the initial radius.

Answer 1

In Bohr's model of the hydrogen atom, the radius of the orbit of the electron in the \( n \)-th state is given by the formula: \[ r_n = n^2 \cdot r_1 \] where: - \( r_n \) is the radius of the orbit in the \( n \)-th state, - \( r_1 \) is the radius of the orbit in the \( n = 1 \) state (also called the Bohr radius), - \( n \) is the principal quantum number. The radius of the orbit for \( n = 3 \) is: \[ r_3 = 3^2 \cdot r_1 = 9r_1 \] The radius of the orbit for \( n = 2 \) is: \[ r_2 = 2^2 \cdot r_1 = 4r_1 \] Now, to find the percentage change in the radius when the electron makes a transition from \( n = 3 \) to \( n = 2 \), we calculate the difference in radius and divide by the initial radius: \[ \text{Percentage change} = \frac{r_3 - r_2}{r_3} \times 100 \] Substitute the values: \[ \text{Percentage change} = \frac{9r_1 - 4r_1}{9r_1} \times 100 = \frac{5r_1}{9r_1} \times 100 = \frac{5}{9} \times 100 \] Thus, the percentage change in the radius is: \[ \text{Percentage change} = 55.56\% \] Hence, the percentage change in the radius of the orbit when the electron makes a transition from \( n = 3 \) to \( n = 2 \) is approximately \( 55.56\% \).