It is teming to think that all possible transituion are permissible and that an atomic spectrum series from the transition of an electron from any intial orbital to any other .However this is not so because a photon a photon has as intrinsic spin angular momentum of `sqrt(2) h// 2pi` corresponding to `S = 1` although it has no charge and no rest mass
On the other hand , an electron has got two typwe of agular momentum orbit angular momentum
`L = [sqrt(l(l+1))] h//2pi`,and spin angular momentum `L_(1) = sqrt(s(s + 1)) h//2pi` arising from orbital motion and spin motion of the electronn during any electton transition must compentum for the angular momentum carried away by the photon .To salary this condition the different between the azisition quantum number of teh orbital witjhin which the transition `(l = 2)` cannot make a transition into as x-orbital `(l = 0)`because the photon cannot carry away enough angular momentum
The maximum orbital angular momentum of an electon with `n = 5` is

Options

A. `sqrt(6) (h)/(2pi)`
B. `sqrt(12) (h)/(2pi)`
C. `sqrt(42) (h)/(2pi)`
D. `sqrt(20) (h)/(2pi)`

Explanation

To solve the question regarding the maximum orbital angular momentum of an electron with a principal quantum number \( n = 5 \), we can follow these steps:

### Step 1: Understand the relationship between quantum numbers and angular momentum
The orbital angular momentum \( L \) of an electron is given by the formula:
\[ L = \sqrt{l(l + 1)} \frac{h}{2\pi} \]
where \( l \) is the azimuthal quantum number, and \( h \) is Planck's constant.

### Step 2: Determine the possible values of \( l \)
For a given principal quantum number \( n \), the azimuthal quantum number \( l \) can take values from \( 0 \) to \( n - 1 \). Therefore, for \( n = 5 \):
\[ l = 0, 1, 2, 3, 4 \]

### Step 3: Identify the maximum value of \( l \)
To find the maximum orbital angular momentum, we need to select the maximum value of \( l \). The maximum value of \( l \) when \( n = 5 \) is:
\[ l = 4 \]

### Step 4: Substitute \( l \) into the angular momentum formula
Now, we substitute \( l = 4 \) into the formula for \( L \):
\[ L = \sqrt{4(4 + 1)} \frac{h}{2\pi} \]

### Step 5: Calculate the value inside the square root
Calculating the expression inside the square root:
\[ 4(4 + 1) = 4 \times 5 = 20 \]

### Step 6: Write the final expression for \( L \)
Now, substituting back into the formula:
\[ L = \sqrt{20} \frac{h}{2\pi} \]

### Conclusion
Thus, the maximum orbital angular momentum of an electron with \( n = 5 \) is:
\[ L = \sqrt{20} \frac{h}{2\pi} \]