According to Bohr's theory, the moment of momentum of an electron revolving in the 4th orbit of a hydrogen atom is:
- \( \frac{8h}{\pi} \)
- \( \frac{h}{\pi} \)
- \( \frac{2h}{\pi} \)
- \( \frac{h}{2\pi} \)
The Correct Option is C
Approach Solution - 1
According to Bohr’s theory, the angular momentum (moment of momentum) \( L \) of an electron in the \( n \)-th orbit is quantized and given by:
\[ L = \frac{n h}{2\pi}, \] where \( h \) is Planck’s constant and \( n \) is the orbit number.
For an electron in the 4th orbit (\( n = 4 \)):
\[ L = \frac{4h}{2\pi} = \frac{2h}{\pi}. \]
Answer: \(\frac{2h}{\pi}\)
Approach Solution -2
According to Bohr’s atomic model, the angular momentum (moment of momentum) of an electron in a stationary orbit is quantized and given by the relation:
\[ mvr = \frac{nh}{2\pi} \] where:
\(m\) = mass of the electron,
\(v\) = velocity of the electron,
\(r\) = radius of the orbit,
\(h\) = Planck’s constant,
\(n\) = principal quantum number (orbit number).
Step 2: Substitute for the 4th orbit
For the 4th orbit, \(n = 4\). Substituting in Bohr’s formula gives:
\[ mvr = \frac{4h}{2\pi} \] Simplify this expression:
\[ mvr = \frac{2h}{\pi} \]
Step 3: Interpret the result
The quantity \(mvr\) represents the moment of momentum or angular momentum of the electron in the orbit. It depends only on the quantum number \(n\) and Planck’s constant \(h\). For the 4th orbit, its value is exactly four times that of the first orbit since \(mvr \propto n\).
Step 4: Final Answer
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