The shortest wavelength of Lyman series of the hydrogen atom is equal to the shortest wavelength of Balmer series of a hydrogen -like atom of atomic number Z. The value of Z is equal to

Options

A. 2
B. 3
C. 4
D. 6

Answer

To solve the problem, we need to find the value of \( Z \) such that the shortest wavelength of the Lyman series of hydrogen is equal to the shortest wavelength of the Balmer series of a hydrogen-like atom with atomic number \( Z \).

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions from higher energy levels down to the first energy level (n=1). The shortest wavelength in the Lyman series occurs when the transition is from \( n=2 \) to \( n=1 \).

Using the Rydberg formula for hydrogen:
\[ \frac{1}{\lambda} = R_H \left( 1 - \frac{1}{2^2} \right) \]
where \( R_H \) is the Rydberg constant for hydrogen.

Calculating this gives:
\[ \frac{1}{\lambda} = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \]

### Step 2: Understand the Balmer Series
The Balmer series corresponds to transitions from higher energy levels down to the second energy level (n=2). The shortest wavelength in the Balmer series occurs when the transition is from \( n=3 \) to \( n=2 \).

Using the Rydberg formula for a hydrogen-like atom with atomic number \( Z \):
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating this gives:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{4} - \frac{1}{9} \right) = R_H Z^2 \left( \frac{9 - 4}{36} \right) = R_H Z^2 \left( \frac{5}{36} \right) \]

### Step 3: Set the Two Wavelength Equations Equal
Since the shortest wavelengths are equal, we can set the two equations equal to each other:
\[ R_H \left( \frac{3}{4} \right) = R_H Z^2 \left( \frac{5}{36} \right) \]

### Step 4: Cancel \( R_H \) and Rearrange
Cancelling \( R_H \) from both sides:
\[ \frac{3}{4} = Z^2 \left( \frac{5}{36} \right) \]

Rearranging gives:
\[ Z^2 = \frac{3}{4} \cdot \frac{36}{5} = \frac{108}{20} = \frac{27}{5} \]

### Step 5: Solve for \( Z \)
Taking the square root of both sides:
\[ Z = \sqrt{\frac{27}{5}} = \frac{3\sqrt{3}}{\sqrt{5}} \approx 2.32 \]

### Conclusion
The value of \( Z \) is approximately \( 2.32 \).