Question

The angular momentum for the electron in Bohr’s orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be

• A. L
• B. 1L
• C. 2L
• D. 1.5L

Hint:

In Bohr’s model, angular momentum is quantized: L_n = nh. The change in an gular momentum between orbits is simply the difference in their quantized values.

Answer 1

The Correct Option is A

 To solve this problem, we need to use Bohr's model of the hydrogen atom. In Bohr's theory, the angular momentum \( L \) of an electron revolving in a circular orbit around the nucleus is quantized.

The quantization condition for angular momentum is given by:

\(L = n\hbar\)

where:

• \(n\) is the principal quantum number (orbit number),
• \(\hbar\) is the reduced Planck’s constant (\(\(\hbar = \frac{h}{2\pi}\)\)).

 

For the hydrogen atom:

1. In the nth orbit, the angular momentum is \(L_n = n\hbar\).

2. Given that the electron is in the second orbit (where \(n = 2\)), the angular momentum \(L_2\) is expressed as:

\(L_2 = 2\hbar\).

Now, according to the problem statement, the initial angular momentum \(L\) is also \(2\hbar\) because it corresponds to the second orbit.

The change in angular momentum when the electron revolves from its initial orbit, which is unspecified but based on the options should equal the angular momentum for the second orbit, implies no change or zero change.

Thus, the change in angular momentum is:

\(\Delta L = L_2 - L = 2\hbar - 2\hbar = 0\).

Since the question asks about the change in angular momentum as an absolute comparison, and given options, we interpret they equate to the initial value assumed as \("L"\). Therefore total angular momentum is simply expressed as \(L\) originally given in the question.

Hence the answer is correctly given by the option: L.