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.

Answer 2

We are given the angular momentum (L) of an electron in a hydrogen atom: \[ L = \frac{3h}{2\pi} \] where \( h \) is Planck's constant. We need to find the kinetic energy (K.E.) of this electron.

Applying Bohr's Model

According to Bohr's postulates for the hydrogen atom, the angular momentum of an electron orbiting the nucleus is quantized and given by the formula: \[ L = n \frac{h}{2\pi} \] where \( n \) is the principal quantum number, representing the orbit number (n = 1, 2, 3, ...).

Comparing the given angular momentum with Bohr's quantization condition: \[ n \frac{h}{2\pi} = \frac{3h}{2\pi} \] This directly implies that the electron is in the orbit with the principal quantum number \( n = 3 \).

Calculating Kinetic Energy

The total energy (E_n) of an electron in the n^th orbit of a hydrogen atom is given by: \[ E_n = -\frac{13.6}{n^2} \text{ eV} \]

In the Bohr model, the kinetic energy (K.E.) of the electron is equal to the negative of its total energy: \[ K.E._n = -E_n \]

Substituting \( n = 3 \) into the formula for total energy: \[ E_3 = -\frac{13.6}{3^2} \text{ eV} = -\frac{13.6}{9} \text{ eV} \] \[ E_3 \approx -1.51 \text{ eV} \]

Now, calculate the kinetic energy for \( n = 3 \): \[ K.E._3 = -E_3 = - \left( -\frac{13.6}{9} \right) \text{ eV} \] \[ K.E._3 = \frac{13.6}{9} \text{ eV} \approx 1.51 \text{ eV} \]

The final answer is ${1.51 \text{ eV}}$ (Option B)