Question

Angular momentum of an electron in hydrogen atom is \(\frac{3h}{2\pi}\) (h is the Planck’s constant). The K.E. of the electron is

• A. 4.35 eV
• B. 1.51 eV
• C. 3.4 eV
• D. 6.8 eV

Answer 1

The Correct Option is B

The angular momentum \(L\) of an electron in the hydrogen atom is given as: \[ L = \frac{3h}{2\pi} \] This represents the third allowed orbit in the Bohr model. According to Bohr's model, the kinetic energy \(K.E.\) of an electron in a hydrogen atom is related to its angular momentum \(L\) by the formula: \[ K.E. = \frac{L^2}{2m r^2} \] But we can also use the formula for the energy levels of the hydrogen atom, which are given by: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] For \(n = 3\) (since the given angular momentum corresponds to the third orbit), we have: \[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \, \text{eV} \] The kinetic energy is the negative of the total energy in this case (since the potential energy is double the kinetic energy and both energies are negative), so: \[ K.E. = 1.51 \, \text{eV} \]

The correct answer is (B) : 1.51 eV.