Ligth emitted during the de excitation of electrons from n=3 to n=2, when incident on metal, photoelectrons following de excitations photoelectrons are just emitted from that metal. In which of the following de excitations photoelectric effect, is not possible ?

Options

A. From n=2 to n=1
B. From n=3 to n=1
C. From n=5 to n=2
D. From n= to n=1

Explanation

To solve the problem, we need to analyze the energy transitions of electrons in an atom and determine whether the emitted light can cause the photoelectric effect in a metal. The key steps are as follows:

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect occurs when light of suitable frequency strikes a metal surface, causing the emission of electrons. The energy of the incident light must be greater than or equal to the work function of the metal.

2. **Calculate the Energy for the Transition from n=3 to n=2**:
The energy emitted during the transition from n=3 to n=2 can be calculated using the formula:
\[ E = E_0 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 2 \) and \( n_2 = 3 \):
\[ E = E_0 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = E_0 \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ E = E_0 \left( \frac{9}{36} - \frac{4}{36} \right) = E_0 \left( \frac{5}{36} \right) \approx 0.14 E_0 \]

3. **Analyze Each Option**:
We will calculate the energy for each transition given in the options and compare it to the minimum energy (0.14 E_0).

- **Option A: n=2 to n=1**:
\[ E = E_0 \left( 1 - \frac{1}{4} \right) = E_0 \left( \frac{3}{4} \right) = 0.75 E_0 \quad (\text{greater than } 0.14 E_0) \]

- **Option B: n=3 to n=1**:
\[ E = E_0 \left( 1 - \frac{1}{9} \right) = E_0 \left( \frac{8}{9} \right) \approx 0.89 E_0 \quad (\text{greater than } 0.14 E_0) \]

- **Option C: n=5 to n=2**:
\[ E = E_0 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = E_0 \left( \frac{1}{4} - \frac{1}{25} \right) \]
Finding a common denominator (100):
\[ E = E_0 \left( \frac{25}{100} - \frac{4}{100} \right) = E_0 \left( \frac{21}{100} \right) = 0.21 E_0 \quad (\text{greater than } 0.14 E_0) \]

- **Option D: n=4 to n=3**:
\[ E = E_0 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = E_0 \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ E = E_0 \left( \frac{16}{144} - \frac{9}{144} \right) = E_0 \left( \frac{7}{144} \right) \approx 0.048 E_0 \quad (\text{less than } 0.14 E_0) \]

4. **Conclusion**:
Since the energy for the transition from n=4 to n=3 is less than the minimum energy required (0.14 E_0), the photoelectric effect is not possible for this transition. Therefore, the correct answer is **Option D**.