The ionization enegry of the electron in the hydrogen atom in its ground state is `13.6 ev`. The atoms are excited to higher energy levels to emit radiations of `6` wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

Options

A. n=3 and n=2 states
B. n=3 and n=1 state
C. n=2 and n=1 state
D. n=4 and n=3 states

Explanation

To solve the problem, we need to determine the maximum wavelength of emitted radiation corresponding to the transition between different energy levels in a hydrogen atom. Here's the step-by-step solution:

### Step 1: Understand Ionization Energy
The ionization energy of the electron in the hydrogen atom in its ground state is given as 13.6 eV. This means that the energy of the electron in the ground state (n=1) is -13.6 eV.

### Step 2: Determine Energy Levels
For hydrogen, the energy levels can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

- For \( n=1 \):
\[ E_1 = -13.6 \, \text{eV} \]
- For \( n=2 \):
\[ E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV} \]
- For \( n=3 \):
\[ E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV} \]
- For \( n=4 \):
\[ E_4 = -\frac{13.6}{4^2} = -0.85 \, \text{eV} \]

### Step 3: Identify Possible Transitions
The problem states that there are six wavelengths emitted, which implies that the electron has been excited to a higher energy level. The maximum principal quantum number that allows for six transitions can be determined from the formula for the number of spectral lines:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
Setting this equal to 6 gives:
\[ n(n-1) = 12 \]
The solution to this is \( n = 4 \). Thus, the electron is excited to \( n = 4 \).

### Step 4: Calculate Energy Differences for Transitions
To find the maximum wavelength, we need to calculate the energy differences for the possible transitions:

1. **Transition from \( n=4 \) to \( n=3 \)**:
\[ E_{4 \to 3} = E_4 - E_3 = (-0.85) - (-1.51) = 0.66 \, \text{eV} \]

2. **Transition from \( n=4 \) to \( n=2 \)**:
\[ E_{4 \to 2} = E_4 - E_2 = (-0.85) - (-3.4) = 2.55 \, \text{eV} \]

3. **Transition from \( n=4 \) to \( n=1 \)**:
\[ E_{4 \to 1} = E_4 - E_1 = (-0.85) - (-13.6) = 12.75 \, \text{eV} \]

4. **Transition from \( n=3 \) to \( n=2 \)**:
\[ E_{3 \to 2} = E_3 - E_2 = (-1.51) - (-3.4) = 1.89 \, \text{eV} \]

5. **Transition from \( n=3 \) to \( n=1 \)**:
\[ E_{3 \to 1} = E_3 - E_1 = (-1.51) - (-13.6) = 12.09 \, \text{eV} \]

6. **Transition from \( n=2 \) to \( n=1 \)**:
\[ E_{2 \to 1} = E_2 - E_1 = (-3.4) - (-13.6) = 10.2 \, \text{eV} \]

### Step 5: Identify the Maximum Wavelength
The maximum wavelength corresponds to the minimum energy difference. From the calculations above, the smallest energy difference is:
\[ E_{4 \to 3} = 0.66 \, \text{eV} \]

### Step 6: Conclusion
Thus, the maximum wavelength of emitted radiation corresponds to the transition between \( n=4 \) and \( n=3 \).

### Final Answer
The maximum wavelength of emitted radiation corresponds to the transition between \( n=4 \) and \( n=3 \).

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