Question

In Bohr atom model, the total energy of the electron in hydrogen atom is -3.4eV. Then its angular momentum about the nucleus of the atom is (h=Planck's constant)

• A. \(\frac{h}{\pi}\)
• B. \(\frac{h}{2\pi}\)
• C. \(\frac{2h}{\pi}\)
• D. \(\frac{4h}{\pi}\)
• E. \(\frac{h}{4\pi}\)

Answer 1

The Correct Option is A

Recall the Bohr Model Energy Levels for Hydrogen:
According to the Bohr model, the energy of an electron in the \(n^{th}\) orbit of a hydrogen atom is given by the formula: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] where \(n\) is the principal quantum number (\(n = 1, 2, 3, ...\)).

Determine the Principal Quantum Number (n):
We are given that the total energy of the electron is \(E = -3.4\) eV. We can use the energy formula to find the value of \(n\) corresponding to this energy level: \[ -3.4 \, \text{eV} = -\frac{13.6}{n^2} \, \text{eV} \] Divide both sides by -1 eV: \[ 3.4 = \frac{13.6}{n^2} \] Solve for \(n^2\): \[ n^2 = \frac{13.6}{3.4} \] \[ n^2 = 4 \] Take the square root (since \(n\) must be positive): \[ n = 2 \] So, the electron is in the second energy level (n=2).

Recall Bohr's Angular Momentum Quantization Condition:
One of the postulates of the Bohr model is that the angular momentum \(L\) of an electron orbiting the nucleus is quantized and is given by: \[ L_n = n \frac{h}{2\pi} \] where \(n\) is the principal quantum number and \(h\) is Planck's constant.

Calculate the Angular Momentum:
Substitute the value \(n=2\) into the angular momentum formula: \[ L_2 = 2 \times \frac{h}{2\pi} \] Simplify the expression: \[ L_2 = \frac{2h}{2\pi} = \frac{h}{\pi} \]

The angular momentum of the electron in the state with energy -3.4 eV (which corresponds to n=2) is \( \frac{h}{\pi} \).

So, the correct option (A): \(\frac h \pi\)

Answer 2

Given: 

• Total energy (E) = -3.4 eV
• This energy corresponds to the n = 2 orbit in Bohr’s hydrogen atom model.

In Bohr's model, the angular momentum of the electron in the n^th orbit is given by: 

\( L = n \cdot \frac{h}{2\pi} \) 

For n = 2: 
\( L = 2 \cdot \frac{h}{2\pi} = \frac{h}{\pi} \) 

✅ Correct Answer: \(\frac{h}{\pi}\)