Question

The wavelength of electron in the third orbit of hydrogen atom is \(6\pi a_0\). The kinetic energy of electron (in J) is equal to \[ \left( \text{Where, } K=\frac{h^2}{\pi^2a_0^2m_e}; \; a_0=\text{radius of first orbit of hydrogen}; \; m_e=\text{mass of electron}, \; h=\text{Planck's constant} \right) \]

• A. \(\dfrac{K}{36}\)
• B. \(\dfrac{K}{108}\)
• C. \(\dfrac{K}{72}\)
• D. \(\dfrac{K}{18}\)

Hint:

For a particle, \[ \boxed{ E_k=\frac{h^2}{2m\lambda^2}. } \] Substitute the given de Broglie wavelength directly and simplify using the given constant.

Answer 1

The Correct Option is C

Step 1: Use the de Broglie relation. The kinetic energy of an electron is \[ E_k = \frac{p^2}{2m_e} = \frac{h^2}{2m_e\lambda^2}. \] Given, \[ \lambda = 6\pi a_0. \]

Step 2: Substitute the wavelength. Therefore, \[ E_k = \frac{h^2} {2m_e(6\pi a_0)^2} = \frac{h^2} {72\pi^2a_0^2m_e}. \] Since \[ K=\frac{h^2}{\pi^2a_0^2m_e}, \] we obtain \[ E_k = \frac{K}{72}. \] Hence, \[ \boxed{E_k=\frac{K}{72}.} \] Therefore, the correct option is \(\boxed{(C)}\).