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

Options

A. Hydrogen atom
B. Deuterium atom
C. single ionized helium
D. Doubly ionized lithium

Answer

To determine in which system the radius of the first orbit (n = 1) will be minimum, we can use the formula for the radius of the nth orbit in a hydrogen-like atom:

\[ r_n = \frac{n^2}{Z} \cdot r_0 \]

where:
- \( r_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number (in this case, \( n = 1 \)),
- \( Z \) is the atomic number,
- \( r_0 \) is a constant (the Bohr radius).

Since we are looking for the minimum radius at \( n = 1 \), we can simplify the formula to:

\[ r_1 = \frac{1^2}{Z} \cdot r_0 = \frac{r_0}{Z} \]

From this equation, we can see that the radius \( r_1 \) is inversely proportional to the atomic number \( Z \). This means that as \( Z \) increases, \( r_1 \) decreases.

Now, let's analyze the options provided:

1. **Hydrogen (H)**: \( Z = 1 \)
2. **Deuterium (D)**: \( Z = 1 \)
3. **Ionized Helium (He\(^+\))**: \( Z = 2 \)
4. **Doubly Ionized Lithium (Li\(^{2+}\))**: \( Z = 3 \)

Now, we can calculate the radius for each option:

- For Hydrogen and Deuterium:
\[ r_1 = \frac{r_0}{1} = r_0 \]

- For Ionized Helium:
\[ r_1 = \frac{r_0}{2} \]

- For Doubly Ionized Lithium:
\[ r_1 = \frac{r_0}{3} \]

From these calculations, we can see that:

- Hydrogen and Deuterium have the same radius \( r_0 \).
- Ionized Helium has a radius of \( \frac{r_0}{2} \).
- Doubly Ionized Lithium has a radius of \( \frac{r_0}{3} \).

Since \( \frac{r_0}{3} < \frac{r_0}{2} < r_0 \), the minimum radius occurs in the system with the highest atomic number, which is the doubly ionized lithium (Li\(^{2+}\)).

**Final Answer**: The radius of the first orbit (n = 1) will be minimum in the system of doubly ionized lithium (Li\(^{2+}\)).