In the above question, if the scientist continues taking data at higher photon energies he will find the next major ''dip'' in the intensity graph at what photon energy ?

Options

A. `(1)/(9) E_(0)`
B. `(8)/(9) E_(0)`
C. `3 E_(0)`
D. `9 E_(0)`

Explanation

To find the next major dip in the intensity graph at higher photon energies, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Dip**:
- The first major dip in the intensity graph occurs due to the transition of electrons in a hydrogen atom from the ground state (n=1) to the first excited state (n=2). The corresponding photon energy for this transition is approximately 1.6 eV.

2. **Identifying the Next Transition**:
- After the transition from n=1 to n=2, the next major dip will occur when the electron transitions from n=1 to n=3. This is because the electrons in the hydrogen atom can only occupy discrete energy levels.

3. **Calculating the Energy Levels**:
- The energy levels of the hydrogen atom are given by the formula:
\[ E_n = -\frac{E_0}{n^2} \]
where \(E_0\) is the energy of the ground state (n=1) and \(n\) is the principal quantum number.

4. **Finding the Energy for n=1 and n=3**:
- For n=1 (ground state):
\[ E_1 = -E_0 \]
- For n=3 (first excited state):
\[ E_3 = -\frac{E_0}{3^2} = -\frac{E_0}{9} \]

5. **Calculating the Energy Difference**:
- The energy difference (ΔE) between the two states (n=1 and n=3) is:
\[ \Delta E = E_3 - E_1 = \left(-\frac{E_0}{9}\right) - \left(-E_0\right) = -\frac{E_0}{9} + E_0 \]
- Simplifying this gives:
\[ \Delta E = E_0 \left(1 - \frac{1}{9}\right) = E_0 \left(\frac{8}{9}\right) \]

6. **Conclusion**:
- Therefore, the photon energy corresponding to the next major dip in the intensity graph (from n=1 to n=3) is:
\[ E_{\text{photon}} = \frac{8E_0}{9} \]

### Final Answer:
The next major dip in the intensity graph will occur at a photon energy of \(\frac{8E_0}{9}\).