QJEEAdv08Ph92e9: JEE Advanced 2008 Physics - Class 12 JEEAdv08Ph92e9 Solution: The ratio of the kinetic energy of the electron for the H atom…

To determine the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He $^+$ ion, we need to consider the Bohr model of the atom. According to the Bohr model, the kinetic energy (KE) of an electron in a hydrogen-like ion in the nth orbit can be given by:

$KE_n = \frac{1}{2} m v_n^2 = \frac{Z^2 e^4 m}{8 \epsilon_0^2 h^2 n^2}$

where:

• $m$ is the mass of the electron
• $Z$ is the atomic number of the nucleus
• $e$ is the charge of the electron
• $\epsilon_0$ is the permittivity of free space
• $h$ is Planck's constant
• $n$ is the principle quantum number, which is 2 in this case

For a hydrogen atom (H) ( $Z = 1$ ) in the $n = 2$ state:

$KE_{H (n=2)} = \frac{1^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{e^4 m}{32 \epsilon_0^2 h^2}$

For a singly ionized helium ion (He $^+$ ) ( $Z = 2$ ) in the $n = 2$ state:

$KE_{He^+ (n=2)} = \frac{2^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{4 e^4 m}{32 \epsilon_0^2 h^2} = \frac{e^4 m}{8 \epsilon_0^2 h^2}$

To find the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He $^+$ ion, we divide the kinetic energies:

$\text{Ratio} = \frac{KE_{H (n=2)}}{KE_{He^+ (n=2)}} = \frac{\frac{e^4 m}{32 \epsilon_0^2 h^2}}{\frac{e^4 m}{8 \epsilon_0^2 h^2}} = \frac{1}{4}$

Therefore, the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He $^+$ ion is:

Option A: $\frac{1}{4}$