Wavelength of radiation emitted when an electron jumps from a state A to state C is 2000 `Å` and it is 6000 Å when the electron jumps from state B to State C, wavelength of the radiation emitted when an electron jumps from state A to B will be

Options

A. `2000 Å`
B. `3000 Å`
C. `4000 Å`
D. `6000 Å`

Answer

To find the wavelength of the radiation emitted when an electron jumps from state A to state B, we can use the information given about the transitions from A to C and B to C.

### Step-by-step Solution:

1. **Identify the given wavelengths:**
- Wavelength when the electron jumps from state A to state C: \( \lambda_{AC} = 2000 \, \text{Å} \)
- Wavelength when the electron jumps from state B to state C: \( \lambda_{BC} = 6000 \, \text{Å} \)

2. **Understand the energy levels:**
- Let \( E_A \), \( E_B \), and \( E_C \) be the energies of states A, B, and C respectively.
- Since \( A \) is at a higher energy level than \( B \) and \( C \), we can conclude that \( E_A > E_B > E_C \).

3. **Use the formula for energy and wavelength:**
- The energy difference corresponding to a transition can be expressed as:
\[ E = \frac{hc}{\lambda} \]
- Therefore, for the transitions:
- From A to C:
\[ E_A - E_C = \frac{hc}{\lambda_{AC}} = \frac{hc}{2000} \]
- From B to C:
\[ E_B - E_C = \frac{hc}{\lambda_{BC}} = \frac{hc}{6000} \]

4. **Set up the equations:**
- From the above, we can write:
\[ E_A - E_C = \frac{hc}{2000} \quad \text{(1)} \]
\[ E_B - E_C = \frac{hc}{6000} \quad \text{(2)} \]

5. **Find the energy difference between states A and B:**
- The difference in energy between states A and B can be expressed as:
\[ E_A - E_B = (E_A - E_C) - (E_B - E_C) \]
- Substituting equations (1) and (2):
\[ E_A - E_B = \left(\frac{hc}{2000}\right) - \left(\frac{hc}{6000}\right) \]

6. **Calculate the right-hand side:**
- To combine the fractions:
\[ E_A - E_B = hc \left(\frac{1}{2000} - \frac{1}{6000}\right) \]
- Finding a common denominator (6000):
\[ E_A - E_B = hc \left(\frac{3 - 1}{6000}\right) = \frac{2hc}{6000} = \frac{hc}{3000} \]

7. **Relate the energy difference to wavelength:**
- Now, using the energy-wavelength relation:
\[ E_A - E_B = \frac{hc}{\lambda_{AB}} \]
- Setting this equal to the previous result:
\[ \frac{hc}{\lambda_{AB}} = \frac{hc}{3000} \]

8. **Solve for the wavelength \( \lambda_{AB} \):**
- Cancel \( hc \) from both sides:
\[ \frac{1}{\lambda_{AB}} = \frac{1}{3000} \]
- Therefore, we find:
\[ \lambda_{AB} = 3000 \, \text{Å} \]

### Final Answer:
The wavelength of the radiation emitted when an electron jumps from state A to state B is \( \lambda_{AB} = 3000 \, \text{Å} \).

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