The radius of the shortest orbit in a one electron system is 18pm it may be

Options

A. Hydrogen
B. deuterium
C. `He^(+)`
D. `Li^(++)`

Answer

To solve the problem of determining which one-electron system has a shortest orbit radius of 18 picometers, we will use the formula for the radius of the nth orbit in a hydrogen-like atom:

\[ R_n = \frac{n^2 \cdot a_0}{Z} \]

where:
- \( R_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number (for the shortest orbit, \( n = 1 \)),
- \( a_0 \) is the Bohr radius (approximately 53 picometers for hydrogen),
- \( Z \) is the atomic number of the element.

### Step-by-Step Solution:

1. **Identify the given radius**:
- The radius of the shortest orbit is given as \( R = 18 \) picometers.

2. **Calculate the radius for hydrogen (Z = 1)**:
- For hydrogen, \( n = 1 \) and \( Z = 1 \):
\[ R_H = \frac{1^2 \cdot 53 \, \text{pm}}{1} = 53 \, \text{pm} \]
- This is not equal to 18 pm.

3. **Calculate the radius for deuterium (Z = 1)**:
- For deuterium, \( n = 1 \) and \( Z = 1 \):
\[ R_D = \frac{1^2 \cdot 53 \, \text{pm}}{1} = 53 \, \text{pm} \]
- This is also not equal to 18 pm.

4. **Calculate the radius for helium ion (He\(^+\), Z = 2)**:
- For helium ion, \( n = 1 \) and \( Z = 2 \):
\[ R_{He} = \frac{1^2 \cdot 53 \, \text{pm}}{2} = \frac{53 \, \text{pm}}{2} = 26.5 \, \text{pm} \]
- This is not equal to 18 pm.

5. **Calculate the radius for lithium ion (Li\(^+\), Z = 3)**:
- For lithium ion, \( n = 1 \) and \( Z = 3 \):
\[ R_{Li} = \frac{1^2 \cdot 53 \, \text{pm}}{3} = \frac{53 \, \text{pm}}{3} \approx 17.67 \, \text{pm} \]
- This is approximately equal to 18 pm.

6. **Conclusion**:
- The only option that gives a radius close to 18 pm is for the lithium ion (Li\(^+\)). Therefore, the answer is that the one-electron system with a shortest orbit radius of 18 pm is the lithium ion.

### Final Answer:
The one-electron system with a shortest orbit radius of 18 pm is the lithium ion (Li\(^+\)).