Question

The angular momentum of an electron in a hydrogen atom is proportional to: (Where \( r \) is the radius of the orbit of the electron)

• A. \( \sqrt{r} \)
• B. \( \frac{1}{r} \)
• C. \( r \)
• D. \( \frac{1}{\sqrt{r}} \)

Answer 1

The Correct Option is A

According to Bohr’s model of the hydrogen atom, the angular momentum \( L \) of an electron in an orbit is quantized and given by:

\( L = n\hbar, \)

where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck’s constant.
For a hydrogen atom, the radius of the \( n \)-th orbit is given by:

\( r_n \propto n^2. \)

Therefore, we can express \( n \) in terms of \( r \):

\( n \propto \sqrt{r}. \)

Substituting this into the expression for angular momentum:

\( L \propto n \propto \sqrt{r}. \)

Hence, the angular momentum of an electron in a hydrogen atom is proportional to \( \sqrt{r}. \)

Answer 2

Step 1: Recall the formula for angular momentum.
For an electron in a hydrogen atom, the angular momentum is given by
L = mvr, where m is the mass of the electron, v is its velocity, and r is the radius of the orbit.

Step 2: Use the centripetal force balance.
The electrostatic force provides the necessary centripetal force for the electron’s circular motion:
k e² / r² = m v² / r
⟹ v² = k e² / (m r)

Step 3: Substitute v into L = m v r.
L = m × √(k e² / (m r)) × r
⟹ L = √(m k e² r)

Step 4: Relation between L and r.
Thus, L ∝ √r.

Final Answer: √r