In which of the following systems will the radius of the first orbit `(n = 1)` be minimum ?

Options

A. Hydrogen atom
B. Deutrium atom
C. Singly ionised helium
D. Doubly ionised lithium

Explanation

To solve the problem of determining which system has the minimum radius of the first orbit (n = 1), we can follow these steps:

### Step 1: Understand the formula for the radius of the orbit
The radius of the nth orbit in a hydrogen-like atom is given by the formula:
\[ r_n = \frac{n^2}{Z} \cdot r_0 \]
where:
- \( r_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number (orbit number),
- \( Z \) is the atomic number (number of protons in the nucleus),
- \( r_0 \) is a constant (approximately 0.529 Å for hydrogen).

### Step 2: Identify the given options and their atomic numbers
We have four options:
1. Hydrogen atom (Z = 1)
2. Deuterium atom (Z = 1, but has one neutron)
3. Singly ionized helium (Z = 2)
4. Doubly ionized lithium (Z = 3)

### Step 3: Calculate the radius for n = 1
Since we are interested in the first orbit (n = 1), we can simplify the formula:
\[ r_1 = \frac{1^2}{Z} \cdot r_0 = \frac{r_0}{Z} \]

### Step 4: Compare the radii based on atomic numbers
Now, we can calculate the radius for each option:
- For Hydrogen (Z = 1):
\[ r_1 = \frac{r_0}{1} = r_0 \]
- For Deuterium (Z = 1):
\[ r_1 = \frac{r_0}{1} = r_0 \]
- For Singly ionized Helium (Z = 2):
\[ r_1 = \frac{r_0}{2} = \frac{r_0}{2} \]
- For Doubly ionized Lithium (Z = 3):
\[ r_1 = \frac{r_0}{3} = \frac{r_0}{3} \]

### Step 5: Determine which radius is the smallest
From the calculations:
- Hydrogen and Deuterium both have a radius of \( r_0 \).
- Singly ionized Helium has a radius of \( \frac{r_0}{2} \).
- Doubly ionized Lithium has a radius of \( \frac{r_0}{3} \).

The smallest radius occurs for Doubly ionized Lithium (Z = 3), which has the minimum radius of \( \frac{r_0}{3} \).

### Conclusion
Thus, the system with the minimum radius of the first orbit (n = 1) is the **doubly ionized lithium** (option 4).

---