A neutron having kinetic energy `E_(0)` collides with singly ionised He atom at rest and move along initial direction. Which of the following statement(s) is/are true for the above mentioned collision.

Options

A. Collision will be perfectly inelastic if `E_(0)=17eV`
B. Collision will be perfectly inelastic is `E_(0)=8.16eV`
C. If `E_(0)=` then ploton of wavelength `lamda=500m` is observed just after the collision if `E_(0)=8.16eV`.
D. Perfectly elastic collision is not possible.

Explanation

To solve the problem, we will analyze the collision between a neutron with kinetic energy \(E_0\) and a singly ionized helium atom at rest. We will apply the principles of conservation of momentum and conservation of energy.

### Step-by-Step Solution:

1. **Identify the masses involved:**
- Let the mass of the neutron be \(m\).
- The mass of the singly ionized helium atom (which has 2 protons and 2 neutrons) is approximately \(4m\).

2. **Initial Conditions:**
- The neutron is moving with an initial velocity \(v_0\) and has kinetic energy \(E_0 = \frac{1}{2}mv_0^2\).
- The helium atom is at rest, so its initial velocity is \(0\).

3. **Conservation of Momentum:**
- Before the collision, the total momentum \(p_i\) is given by:
\[ p_i = mv_0 + 0 = mv_0 \]
- After the collision, let the final velocity of the system be \(v\). The total momentum \(p_f\) after the collision is:
\[ p_f = (m + 4m)v = 5mv \]
- By conservation of momentum:
\[ mv_0 = 5mv \implies v = \frac{v_0}{5} \]

4. **Conservation of Energy:**
- The initial kinetic energy of the neutron is:
\[ KE_i = \frac{1}{2}mv_0^2 \]
- After the collision, the kinetic energy of the system is:
\[ KE_f = \frac{1}{2}(5m)v^2 = \frac{5m}{2}\left(\frac{v_0}{5}\right)^2 = \frac{5m}{2} \cdot \frac{v_0^2}{25} = \frac{mv_0^2}{10} \]
- The change in energy (if any) due to the collision is:
\[ \Delta E = KE_i - KE_f = \frac{1}{2}mv_0^2 - \frac{mv_0^2}{10} = \frac{5mv_0^2}{10} - \frac{mv_0^2}{10} = \frac{4mv_0^2}{10} = \frac{2mv_0^2}{5} \]

5. **Relate \(\Delta E\) to \(E_0\):**
- Since \(E_0 = \frac{1}{2}mv_0^2\), we can express \(\Delta E\) in terms of \(E_0\):
\[ \Delta E = \frac{2}{5}E_0 \]

6. **Analyzing the statements:**
- If \(E_0 = 17 \text{ eV}\), then \(\Delta E = \frac{2}{5} \times 17 = 6.8 \text{ eV}\) (not zero, hence the collision is inelastic).
- If \(E_0 = 8.16 \text{ eV}\), then \(\Delta E = \frac{2}{5} \times 8.16 = 3.264 \text{ eV}\) (again not zero, hence the collision is inelastic).
- The photon emitted has energy equal to \(\Delta E\), which can be checked against the wavelength given in the problem.

7. **Conclusion:**
- The collision is not perfectly elastic since \(\Delta E \neq 0\).
- The statements regarding the energy values and the nature of the collision can be verified based on the calculations above.