Question: A neutron beam, in which each neutron has same kinetic energy, is passed through a sample of hydrogen like gas (but not hydrogen) in ground state. Due to collision of neutrons with the ions of the gas, ions are excited and then they emit photons. Six spectral lines are obtained in which one of the lines is of wavelength $ \text{(}6200/51\text{)}nm $ . What is the minimum possible value of kinetic energy of the neutrons for this to be possible. The mass of neutron and proton can be assumed to be nearly same. Give answer in the form $ 25a\times {{10}^{-2eV}} $ and find value of a.

Solution:

Step 1: Six spectral lines are observed, so $\frac{n(n-1)}{2} = 6$, giving $n = 4$. The ions are excited from the ground state ($n=1$) to $n=4$.

Step 2: Identify the gas. One spectral line has $\lambda = \frac{6200}{51}$ nm $\approx 121.6$ nm.

For a hydrogen-like ion with atomic number $Z$:

$$\frac{1}{\lambda} = R Z^{2}\left(\frac{1}{n _1^{2}} - \frac{1}{n _2^{2}}\right)$$

Testing $Z = 2$ (He$^+$) for the transition $n = 4 \to n = 2$:

$$\frac{1}{\lambda} = R \times 4 \times \left(\frac{1}{4} - \frac{1}{16}\right) = R \times 4 \times \frac{3}{16} = \frac{3R}{4}$$

$$\lambda = \frac{4}{3R} = \frac{4}{3 \times 1.097 \times 10^{7}} \approx 121.5\ \text{nm} = \frac{6200}{51}\ \text{nm} \checkmark$$

So the gas is He$^+$ ($Z = 2$).

Step 3: Energy required to excite He$^+$ from $n = 1$ to $n = 4$:

$$\Delta E = 13.6 \times Z^{2} \times \left(\frac{1}{1^{2}} - \frac{1}{4^{2}}\right) = 13.6 \times 4 \times \frac{15}{16} = 51.0\ \text{eV}$$

Step 4: In the collision, the neutron (mass $m$) hits the He$^+$ ion (mass $M \approx 4m$). Using conservation of momentum and energy, the minimum kinetic energy of the neutron is:

$$KE _{\min} = \Delta E \times \frac{m + M}{M} = 51.0 \times \frac{1 + 4}{4} = 51.0 \times \frac{5}{4} = 63.75\ \text{eV}$$

Step 5: Express in the required form: $63.75\ \text{eV} = 6375 \times 10^{-2}\ \text{eV} = 25 \times 255 \times 10^{-2}\ \text{eV}$.

Therefore $a = 255$.

Answer: 255