Change in angular momentum when an electron makes a transition corresponding to the 3rd line of the Balmer series in `Li^(2+)` ion is :

Options

A. `(h)/(2pi)`
B. `(2h)/(2 pi)`
C. `(3h)/(2pi)`
D. `(4h)/(2pi)`

Answer

To solve the problem of finding the change in angular momentum when an electron makes a transition corresponding to the 3rd line of the Balmer series in the \(Li^{2+}\) ion, we can follow these steps:

### Step 1: Identify the Transition
The 3rd line of the Balmer series corresponds to the transition of an electron from the 5th energy level (n=5) to the 2nd energy level (n=2).

### Step 2: Use the Formula for Angular Momentum
According to Bohr's model, the angular momentum \(L\) of an electron in a hydrogen-like atom is given by:
\[ L = n \frac{h}{2\pi} \]
where \(n\) is the principal quantum number and \(h\) is Planck's constant.

### Step 3: Calculate Initial Angular Momentum
For the initial state (n=5):
\[ L_1 = 5 \frac{h}{2\pi} \]

### Step 4: Calculate Final Angular Momentum
For the final state (n=2):
\[ L_2 = 2 \frac{h}{2\pi} \]

### Step 5: Calculate Change in Angular Momentum
The change in angular momentum \(\Delta L\) is given by:
\[ \Delta L = L_1 - L_2 \]
Substituting the values we calculated:
\[ \Delta L = \left(5 \frac{h}{2\pi}\right) - \left(2 \frac{h}{2\pi}\right) = \left(5 - 2\right) \frac{h}{2\pi} = 3 \frac{h}{2\pi} \]

### Step 6: Conclusion
Thus, the change in angular momentum when an electron transitions corresponding to the 3rd line of the Balmer series in \(Li^{2+}\) ion is:
\[ \Delta L = 3 \frac{h}{2\pi} \]