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 and at rest. 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 (6200/51) nm.Which gas is this?

Options

A. `H`
B. `D`
C. `He^(+)`
D. `Li^(+2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a neutron beam colliding with a hydrogen-like gas in the ground state, which results in the excitation of ions and the emission of photons. We are given that six spectral lines are observed, and one of these lines has a wavelength of \( \frac{6200}{51} \) nm. Our goal is to identify the gas.

### Step 2: Use the Formula for Spectral Lines
The number of spectral lines \( N \) produced by transitions from an excited state \( n \) to lower energy states can be given by the formula:
\[ N = \frac{n(n-1)}{2} \]
Given that \( N = 6 \), we can set up the equation:
\[ \frac{n(n-1)}{2} = 6 \]
Multiplying both sides by 2 gives:
\[ n(n-1) = 12 \]

### Step 3: Solve the Quadratic Equation
Rearranging the equation gives:
\[ n^2 - n - 12 = 0 \]
Now we can factor this quadratic equation:
\[ (n - 4)(n + 3) = 0 \]
This gives us two solutions:
\[ n = 4 \quad \text{and} \quad n = -3 \]
Since \( n \) must be a positive integer, we take \( n = 4 \).

### Step 4: Identify Excited States
The ground state corresponds to \( n = 1 \). The excited states are:
- \( n = 2 \)
- \( n = 3 \)
- \( n = 4 \)

### Step 5: Calculate the Energy Difference
The wavelength of one of the emitted photons is given as \( \frac{6200}{51} \) nm. To find the energy difference corresponding to this wavelength, we use the formula:
\[ \Delta E = \frac{12400}{\lambda} \quad \text{(in eV, where } \lambda \text{ is in angstroms)} \]
First, convert the wavelength from nm to angstroms:
\[ \lambda = \frac{6200}{51} \text{ nm} = \frac{6200 \times 10}{51} \text{ angstroms} \]
Now substituting this into the energy formula:
\[ \Delta E = \frac{12400 \times 51}{62000} = 10.2 \text{ eV} \]

### Step 6: Identify the Gas
Now we need to determine which hydrogen-like ion corresponds to an energy difference of \( 10.2 \) eV for a transition from \( n = 4 \) to \( n = 2 \) or \( n = 1 \).

For hydrogen-like ions, the energy levels are given by:
\[ E_n = -\frac{Z^2 \cdot 13.6}{n^2} \text{ eV} \]
Where \( Z \) is the atomic number.

Calculating the energy difference for \( Z = 2 \) (Helium ion \( He^+ \)):
\[ E_4 = -\frac{2^2 \cdot 13.6}{4} = -\frac{54.4}{4} = -13.6 \text{ eV} \]
\[ E_2 = -\frac{2^2 \cdot 13.6}{2^2} = -\frac{54.4}{4} = -13.6 \text{ eV} \]
The energy difference from \( n = 4 \) to \( n = 2 \) is:
\[ \Delta E = E_2 - E_4 = -\left(-\frac{54.4}{4}\right) - \left(-\frac{54.4}{16}\right) = 10.2 \text{ eV} \]
Thus, the gas is \( He^+ \).

### Final Answer
The gas is \( He^+ \) (Helium ion).

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