Calculate the smaller wavelength of radiation that may be emitted by (a) hydrogen (b)`He^(+)`and (c ) `Li^(++)`

Explanation

To calculate the smaller wavelength of radiation emitted by hydrogen, helium ion (He⁺), and lithium ion (Li²⁺), we can use the formula derived from the Bohr model of the atom:

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

Where:
- \( \lambda \) is the wavelength,
- \( Z \) is the atomic number,
- \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

Since we want to find the smallest wavelength, we will consider transitions from the first energy level (n=1) to infinity (n=∞), which corresponds to the maximum energy transition.

### Step-by-Step Solution:

**(a) For Hydrogen (H):**
1. **Identify parameters:**
- \( Z = 1 \) (for hydrogen),
- \( n_1 = 1 \),
- \( n_2 = \infty \).

2. **Apply the formula:**
\[ \frac{1}{\lambda} = 1^2 \cdot R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
\[ \frac{1}{\lambda} = R \left( 1 - 0 \right) = R \]

3. **Substitute the value of R:**
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1} \]
\[ \lambda = \frac{1}{1.097 \times 10^7} \approx 9.1 \times 10^{-8} \, \text{m} = 91 \, \text{nm} \]

**(b) For Helium Ion (He⁺):**
1. **Identify parameters:**
- \( Z = 2 \) (for He⁺),
- \( n_1 = 1 \),
- \( n_2 = \infty \).

2. **Apply the formula:**
\[ \frac{1}{\lambda} = 2^2 \cdot R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
\[ \frac{1}{\lambda} = 4R \]

3. **Substitute the value of R:**
\[ \frac{1}{\lambda} = 4 \cdot 1.097 \times 10^7 \, \text{m}^{-1} \]
\[ \lambda = \frac{1}{4 \cdot 1.097 \times 10^7} \approx \frac{1}{4.388 \times 10^7} \approx 2.3 \times 10^{-8} \, \text{m} = 23 \, \text{nm} \]

**(c) For Lithium Ion (Li²⁺):**
1. **Identify parameters:**
- \( Z = 3 \) (for Li²⁺),
- \( n_1 = 1 \),
- \( n_2 = \infty \).

2. **Apply the formula:**
\[ \frac{1}{\lambda} = 3^2 \cdot R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
\[ \frac{1}{\lambda} = 9R \]

3. **Substitute the value of R:**
\[ \frac{1}{\lambda} = 9 \cdot 1.097 \times 10^7 \, \text{m}^{-1} \]
\[ \lambda = \frac{1}{9 \cdot 1.097 \times 10^7} \approx \frac{1}{9.873 \times 10^7} \approx 1.01 \times 10^{-8} \, \text{m} = 10 \, \text{nm} \]

### Final Results:
- **Hydrogen (H):** \( \lambda \approx 91 \, \text{nm} \)
- **Helium Ion (He⁺):** \( \lambda \approx 23 \, \text{nm} \)
- **Lithium Ion (Li²⁺):** \( \lambda \approx 10 \, \text{nm} \)