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 to n=1 states
B. n=4 to n=3 states
C. n=3 to n=2 states
D. n=2 to n=1 states

Explanation

To solve the problem, we need to determine the transition that corresponds to the maximum wavelength of emitted radiation when a hydrogen atom transitions between energy levels. Here’s a step-by-step breakdown of the solution:

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

### Step 2: Determine the Number of Wavelengths
We know that the number of spectral lines (wavelengths) emitted when an electron transitions between energy levels can be calculated using the formula:
\[ \text{Number of spectral lines} = \frac{n_2(n_2 - n_1)}{2} \]
where \( n_1 \) is the lower energy level and \( n_2 \) is the upper energy level.

Given that the number of wavelengths is 6, we can set up the equation:
\[ \frac{n_2(n_2 - 1)}{2} = 6 \]
Multiplying both sides by 2 gives:
\[ n_2(n_2 - 1) = 12 \]

### Step 3: Solve for \( n_2 \)
To find \( n_2 \), we can rearrange the equation:
\[ n_2^2 - n_2 - 12 = 0 \]
This is a quadratic equation. We can solve for \( n_2 \) using the quadratic formula:
\[ n_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -1, c = -12 \):
\[ n_2 = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm 7}{2} \]
Calculating the two possible values:
\[ n_2 = \frac{8}{2} = 4 \quad \text{or} \quad n_2 = \frac{-6}{2} = -3 \quad (\text{not valid}) \]
Thus, \( n_2 = 4 \).

### Step 4: Identify Possible Transitions
The possible transitions from \( n_2 = 4 \) to \( n_1 = 1 \) are:
- 4 to 1
- 4 to 3
- 4 to 2
- 3 to 1
- 3 to 2
- 2 to 1

### Step 5: Determine Maximum Wavelength
To find the transition that corresponds to the maximum wavelength, we need to find the transition with the minimum energy difference. The energy levels of the hydrogen atom are given by:
\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]
The energy difference between two levels \( n_i \) and \( n_f \) is:
\[ \Delta E = E_{n_f} - E_{n_i} = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2} \]

### Step 6: Calculate Energy Differences
We will calculate the energy differences for the relevant transitions:
1. **4 to 3**:
\[ \Delta E = -\frac{13.6}{3^2} + \frac{13.6}{4^2} = -\frac{13.6}{9} + \frac{13.6}{16} \]
Calculate the values:
\[ \Delta E = -1.51 + 0.85 = -0.66 \text{ eV} \]

2. **4 to 2**:
\[ \Delta E = -\frac{13.6}{2^2} + \frac{13.6}{4^2} = -\frac{13.6}{4} + \frac{13.6}{16} \]
Calculate the values:
\[ \Delta E = -3.4 + 0.85 = -2.55 \text{ eV} \]

3. **4 to 1**:
\[ \Delta E = -\frac{13.6}{1^2} + \frac{13.6}{4^2} = -13.6 + 0.85 = -12.75 \text{ eV} \]

4. **3 to 2**:
\[ \Delta E = -\frac{13.6}{2^2} + \frac{13.6}{3^2} = -3.4 + 1.51 = -1.89 \text{ eV} \]

5. **3 to 1**:
\[ \Delta E = -\frac{13.6}{1^2} + \frac{13.6}{3^2} = -13.6 + 1.51 = -12.09 \text{ eV} \]

6. **2 to 1**:
\[ \Delta E = -\frac{13.6}{1^2} + \frac{13.6}{2^2} = -13.6 + 3.4 = -10.2 \text{ eV} \]

### Conclusion
The transition with the smallest energy difference (and thus the maximum wavelength) is from **4 to 3**. Therefore, 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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