Calculate Kinetic Energy of Electron in Second Bohr Orbit

To calculate the kinetic energy of an electron residing in the second Bohr orbit of a hydrogen atom, we use Bohr's postulates.

Step 1: Quantization of Angular Momentum According to Bohr's quantization condition, the orbital angular momentum $L$ of an electron in the $n$-th orbit is given by: $m v r_n = \frac{n h}{2\pi}$

where:

• $m$ is the mass of the electron,
• $v$ is the orbital velocity of the electron,
• $r_n$ is the radius of the $n$-th orbit,
• $h$ is Planck's constant,
• $n$ is the principal quantum number.

For the second orbit ($n = 2$): $m v r_2 = \frac{2h}{2\pi} = \frac{h}{\pi}$

Step 2: Radius of the Second Bohr Orbit The radius of the $n$-th orbit in a hydrogen atom ($Z = 1$) is related to the Bohr radius $a_0$ by: $r_n = n^2 a_0$

For $n = 2$: $r_2 = 2^2 a_0 = 4 a_0$

Step 3: Calculating Velocity $v$ Substitute $r_2 = 4 a_0$ into the angular momentum equation: $m v (4 a_0) = \frac{h}{\pi}$

Solving for $v$: $v = \frac{h}{4\pi m a_0}$

Step 4: Calculating Kinetic Energy The kinetic energy ($K.E.$) of the electron is given by: $K.E. = \frac{1}{2} m v^2$

Substituting the value of $v$: $K.E. = \frac{1}{2} m \left( \frac{h}{4\pi m a_0} \right)^2$

$K.E. = \frac{1}{2} m \left( \frac{h^2}{16\pi^2 m^2 a_0^2} \right)$

$K.E. = \frac{h^2}{32\pi^2 m a_0^2}$

Hence, the correct option is C.