Question

The speed of an electron in an orbit of hydrogen atom is \( \frac{1}{274} \) times the speed of light in vacuum. The kinetic energy of the electron in the orbit is:

• A. 3.4 eV
• B. 10.2 eV
• C. 13.6 eV
• D. 5.4 eV

Hint:

The kinetic energy of an electron in the first orbit of a hydrogen atom is \( 3.4 \, \text{eV} \), which is half of the total energy (\(-13.6 \, \text{eV}\)).

Answer 1

The Correct Option is A

The kinetic energy of the electron in the first orbit of hydrogen is given by: \[ K.E. = \frac{1}{2} mv^2 \] From Bohr's model, the total energy of the electron in the first orbit is \( -13.6 \, \text{eV} \). The kinetic energy is half of the total energy, and its positive value is: \[ K.E. = 3.4 \, \text{eV} \] Thus, the kinetic energy of the electron in the first orbit is \( 3.4 \, \text{eV} \).