In which of the following, maximum wavelength is emitted ?

Options

A. 4 to 1 in H
B. 5 to 1in `H^(+)`
C. 6 to 1 in `Li^(+2)`
D. 6 to 1 in `Be^(+3)`

Answer

To determine which of the given transitions emits the maximum wavelength, we can use the formula for the wavelength of emitted light during electronic transitions in hydrogen-like atoms:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \(\lambda\) is the wavelength,
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number,
- \(n_1\) is the lower energy level,
- \(n_2\) is the higher energy level.

The maximum wavelength corresponds to the minimum value of \(\frac{1}{\lambda}\), so we will calculate \(\frac{1}{\lambda}\) for each transition and then find \(\lambda\) for each case.

### Step 1: Calculate for the first option (Hydrogen, \(n_2 = 4\) to \(n_1 = 1\))
- \(Z = 1\)
- \(n_1 = 1\), \(n_2 = 4\)

\[ \frac{1}{\lambda_1} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{15}{16} \right) \]

### Step 2: Calculate for the second option (Helium ion, \(H^+\), \(n_2 = 5\) to \(n_1 = 1\))
- \(Z = 2\)
- \(n_1 = 1\), \(n_2 = 5\)

\[ \frac{1}{\lambda_2} = R \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = R \cdot 4 \left( 1 - \frac{1}{25} \right) = R \cdot 4 \left( \frac{24}{25} \right) \]

### Step 3: Calculate for the third option (Lithium ion, \(Li^{2+}\), \(n_2 = 6\) to \(n_1 = 1\))
- \(Z = 3\)
- \(n_1 = 1\), \(n_2 = 6\)

\[ \frac{1}{\lambda_3} = R \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{6^2} \right) = R \cdot 9 \left( 1 - \frac{1}{36} \right) = R \cdot 9 \left( \frac{35}{36} \right) \]

### Step 4: Calculate for the fourth option (Beryllium ion, \(Be^{3+}\), \(n_2 = 6\) to \(n_1 = 1\))
- \(Z = 4\)
- \(n_1 = 1\), \(n_2 = 6\)

\[ \frac{1}{\lambda_4} = R \cdot 4^2 \left( \frac{1}{1^2} - \frac{1}{6^2} \right) = R \cdot 16 \left( 1 - \frac{1}{36} \right) = R \cdot 16 \left( \frac{35}{36} \right) \]

### Step 5: Compare the values
Now we will compare the expressions for \(\frac{1}{\lambda}\):

1. \(\frac{1}{\lambda_1} = R \cdot \frac{15}{16}\)
2. \(\frac{1}{\lambda_2} = R \cdot \frac{96}{25}\)
3. \(\frac{1}{\lambda_3} = R \cdot \frac{315}{36}\)
4. \(\frac{1}{\lambda_4} = R \cdot \frac{560}{36}\)

To find the maximum wavelength, we need to find the minimum of these values.

### Conclusion
After comparing the fractions, we find that \(\frac{1}{\lambda_1}\) gives the smallest value, indicating that \(\lambda_1\) is the largest. Therefore, the maximum wavelength is emitted in the first case (Hydrogen transition from \(n=4\) to \(n=1\)).

### Final Answer
The maximum wavelength is emitted in the transition from \(n=4\) to \(n=1\) in Hydrogen.

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