Energy of level 1,2,3 of a certain atom corresponds to increasing value of energy `E_(1) lt E_(2) lt E_(3)` . If `lambda_(1) , lambda_(2)` and `lambda_(3)` are the wavelength of radiation corresponding to transition `3 rarr 2, 2 rarr1` and `3 rarr1` respectively. Which of the following statement is `//` are correct ?

Options

A. `(1)/(lambda_(3))= ( 1)/(lambda_(1))+(1)/(lambda_(2))`
B. `lambda_(3)= (lambda_(1) lambda_(2))/(lambda_(1)+lambda_(2))`
C. `(1)/(lambda_(2))= ( 1)/(lambda_(1))+(1)/(lambda_(3))`
D. `lambda_(2)= (lambda_(1) lambda_(3))/(lambda_(1)+lambda_(2))`

Answer

To solve the problem step by step, we need to analyze the transitions between energy levels in the atom and the relationships between energy, wavelength, and frequency.

### Step-by-Step Solution:

1. **Understanding Energy Levels**:
- We are given that the energy levels of the atom are such that \( E_1 < E_2 < E_3 \). This means that \( E_1 \) is the lowest energy state, followed by \( E_2 \), and \( E_3 \) is the highest.

2. **Identifying Transitions**:
- The transitions mentioned are:
- Transition from level 3 to level 2: \( \lambda_1 \)
- Transition from level 2 to level 1: \( \lambda_2 \)
- Transition from level 3 to level 1: \( \lambda_3 \)

3. **Energy and Wavelength Relationship**:
- The energy of a photon emitted or absorbed during a transition can be expressed using the formula:
\[ E = \frac{hc}{\lambda} \]
- Where \( h \) is Planck's constant and \( c \) is the speed of light.

4. **Energy Differences**:
- For the transitions, we can express the energies as:
- For \( 3 \to 2 \): \( E_{3 \to 2} = E_3 - E_2 = \frac{hc}{\lambda_1} \)
- For \( 2 \to 1 \): \( E_{2 \to 1} = E_2 - E_1 = \frac{hc}{\lambda_2} \)
- For \( 3 \to 1 \): \( E_{3 \to 1} = E_3 - E_1 = \frac{hc}{\lambda_3} \)

5. **Adding Energies**:
- According to the principle of conservation of energy, the total energy change for the transition from 3 to 1 should equal the sum of the energies for the transitions from 3 to 2 and 2 to 1:
\[ E_{3 \to 1} = E_{3 \to 2} + E_{2 \to 1} \]
- Substituting the expressions for energy:
\[ \frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} \]

6. **Simplifying the Equation**:
- Dividing through by \( hc \) (which is constant and non-zero):
\[ \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \]
- This can be rearranged to find \( \lambda_3 \):
\[ \frac{1}{\lambda_3} = \frac{\lambda_1 + \lambda_2}{\lambda_1 \lambda_2} \]
- Therefore,
\[ \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} \]

7. **Conclusion**:
- The derived formula shows that \( \lambda_3 \) is indeed equal to \( \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} \). Hence, the correct statement among the options provided corresponds to this relationship.