Electrons having kinetic energy 30 eV are made to collide with atomic hydrogen gas (in ground state) and 42.5% of electron energy is used to excite the hydrogen wavelength in emission spectra are

Options

A. 3
B. 6
C. 12
D. 18

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the energy used to excite the hydrogen atom
The kinetic energy of the electrons is given as 30 eV. We need to find out how much of this energy is used to excite the hydrogen atom.

\[ \text{Energy used} = 42.5\% \text{ of } 30 \text{ eV} = 0.425 \times 30 \text{ eV} = 12.75 \text{ eV} \]

### Step 2: Determine the energy levels of hydrogen
The energy levels of a hydrogen atom can be calculated using the formula:

\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]

Calculating the energy for the first few levels:
- For \( n = 1 \):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6}{2^2} = -3.4 \text{ eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{13.6}{3^2} \approx -1.51 \text{ eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{13.6}{4^2} = -0.85 \text{ eV} \]

### Step 3: Calculate the energy differences for transitions
To find out how far the electron can be excited, we can calculate the energy required to transition from \( n = 1 \) to higher levels.

- From \( n = 1 \) to \( n = 2 \):
\[ \Delta E_{1 \to 2} = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \text{ eV} \]
- From \( n = 1 \) to \( n = 3 \):
\[ \Delta E_{1 \to 3} = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV} \]
- From \( n = 1 \) to \( n = 4 \):
\[ \Delta E_{1 \to 4} = E_4 - E_1 = -0.85 - (-13.6) = 12.75 \text{ eV} \]

### Step 4: Determine possible transitions
Since the energy gained by the hydrogen electron is 12.75 eV, it can transition from \( n = 1 \) to \( n = 4 \).

### Step 5: Calculate the number of unique wavelengths emitted
The number of unique wavelengths emitted during the de-excitation can be calculated using the formula:

\[ \text{Number of unique wavelengths} = \frac{(N_2 - N_1)(N_2 - N_1 + 1)}{2} \]

Here, \( N_2 = 4 \) and \( N_1 = 1 \):

\[ \text{Number of unique wavelengths} = \frac{(4 - 1)(4 - 1 + 1)}{2} = \frac{(3)(4)}{2} = 6 \]

### Final Answer
The number of unique wavelengths emitted is **6**.

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