Question

An electron in Bohr model of hydrogen atom makes a transition from energy level \(-1.51 \, \text{eV}\) to \(-3.40 \, \text{eV}\). Calculate the change in the radius of its orbit. The radius of orbit of electron in its ground state is \(0.53 \, \text{\AA}\).

Hint:

In the Bohr model, the radius of the electron's orbit is proportional to the square of the principal quantum number \( n \).

Answer 1

The electron's orbital radius \( r_n \) in the Bohr model is determined by \( r_n = r_1 n^2 \), where \( r_1 = 0.53 \, \text{\AA} \) represents the ground state orbit radius and \( n \) is the principal quantum number. Energy levels are defined as \( E_n = -\frac{13.6 \, \text{eV}}{n^2} \). For \( E_n = -1.51 \, \text{eV} \), the principal quantum number \( n_1 \) is calculated as: \[ -1.51 = -\frac{13.6}{n_1^2} \implies n_1^2 = \frac{13.6}{1.51} \implies n_1 = 3 \]. For \( E_n = -3.40 \, \text{eV} \), the principal quantum number \( n_2 \) is calculated as: \[ -3.40 = -\frac{13.6}{n_2^2} \implies n_2^2 = \frac{13.6}{3.40} \implies n_2 = 2 \]. The corresponding orbital radii are \( r_{n_1} = r_1 n_1^2 = 0.53 \times 9 = 4.77 \, \text{\AA} \) and \( r_{n_2} = r_1 n_2^2 = 0.53 \times 4 = 2.12 \, \text{\AA} \). The difference in these radii is \( \Delta r = r_{n_1} - r_{n_2} = 4.77 - 2.12 = 2.65 \, \text{\AA} \). Therefore, the change in orbital radius is \( 2.65 \, \text{\AA} \).