Among the following transition in hydrogen and hydrogen-like spectrum, which one emits light of longest wavelength ?

Options

A. `n = 2` "to" `n = 1` "for" `H`
B. `n = 4` "to"` n = 3` "for"` Li^(2+)`
C. `n = 4` "to" `n = 3` "for" `He^(o+)`
D. `n = 5` "to" `n = 2` "for" `H`

Explanation

To determine which transition emits light of the longest wavelength among the given options, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy of a photon is inversely proportional to its wavelength. This means that the longer the wavelength, the lower the energy. The formula is given by:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) = energy of the photon
- \( h \) = Planck's constant
- \( c \) = speed of light
- \( \lambda \) = wavelength

### Step 2: Use the Rydberg formula
The Rydberg formula for hydrogen and hydrogen-like atoms is:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R \) = Rydberg constant
- \( Z \) = atomic number of the atom
- \( n_1 \) and \( n_2 \) = principal quantum numbers (with \( n_2 > n_1 \))

### Step 3: Calculate the wavelength for each transition

1. **For hydrogen (n=1 to n=2)**:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
\[ \lambda = \frac{4}{3R} \quad \text{(or approximately } 1.33R\text{)} \]

2. **For lithium ion (Li\(^{2+}\), n=3 to n=4)**:
\[ \frac{1}{\lambda} = R \cdot 3^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \cdot 9 \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = R \cdot 9 \cdot \frac{7}{144} = \frac{63R}{144} \]
\[ \lambda = \frac{144}{63R} \quad \text{(or approximately } 2.28R\text{)} \]

3. **For helium ion (He\(^{+}\), n=3 to n=4)**:
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \cdot 4 \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = R \cdot 4 \cdot \frac{7}{144} = \frac{28R}{144} \]
\[ \lambda = \frac{144}{28R} \quad \text{(or approximately } 5.14R\text{)} \]

4. **For hydrogen (n=2 to n=5)**:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{4} - \frac{1}{25} \right) \]
\[ = R \cdot \frac{21}{100} \]
\[ \lambda = \frac{100}{21R} \quad \text{(or approximately } 4.76R\text{)} \]

### Step 4: Compare the wavelengths
Now we have the wavelengths for each transition:
- For hydrogen (n=1 to n=2): \( \lambda \approx 1.33R \)
- For lithium ion (n=3 to n=4): \( \lambda \approx 2.28R \)
- For helium ion (n=3 to n=4): \( \lambda \approx 5.14R \)
- For hydrogen (n=2 to n=5): \( \lambda \approx 4.76R \)

The longest wavelength is \( 5.14R \) from the transition of helium ion (He\(^{+}\), n=3 to n=4).

### Final Answer
Thus, the transition that emits light of the longest wavelength is option **3** (n=3 to n=4 for helium unipositive ion).

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