From quantisation of angular momentum one gets for hydrogen atom, the radius of the `n^th` orbit as `r_n=((n^2)/(m_e))((h)/(2pi))^2((4piepsilon_0)/(e^2))`
For a hydrogen like atom of atomic number Z,

Options

A. the radius of the first orbit will be the same
B. `r_n` will be greater for larger Z values
C. `r_n` will be smaller for larger Z values
D. none of these

Explanation

To solve the problem regarding the radius of the nth orbit in a hydrogen-like atom with atomic number Z, we will follow these steps:

### Step 1: Understand the formula for the radius of the nth orbit in a hydrogen atom
The radius of the nth orbit for a hydrogen atom is given by the formula:
\[ r_n = \frac{n^2}{m_e} \left(\frac{h}{2\pi}\right)^2 \left(\frac{4\pi\epsilon_0}{e^2}\right) \]
where:
- \( n \) is the principal quantum number,
- \( m_e \) is the mass of the electron,
- \( h \) is Planck's constant,
- \( \epsilon_0 \) is the permittivity of free space,
- \( e \) is the charge of the electron.

### Step 2: Modify the formula for a hydrogen-like atom
For a hydrogen-like atom with atomic number \( Z \), the radius of the nth orbit can be modified to account for the nuclear charge. The formula becomes:
\[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \]
This indicates that the radius depends on the atomic number \( Z \).

### Step 3: Analyze the dependence on atomic number \( Z \)
From the modified formula, we can see that as \( Z \) increases, the term \( \frac{1}{Z} \) indicates that the radius \( r_n \) will decrease. This means:
- For larger values of \( Z \), the radius of the nth orbit becomes smaller.

### Step 4: Conclusion
Thus, we conclude that for a hydrogen-like atom, the radius of the nth orbit decreases as the atomic number \( Z \) increases. Therefore, the correct answer to the question is that \( r_n \) will be smaller for larger \( Z \) values.

### Final Answer:
The radius \( r_n \) will be smaller for larger \( Z \) values.

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