The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen-like ion. The atomic number of hydrogen-like ion is

• A. 4
• B. 1
• C. 2
• D. 3

The problem involves comparing the wavelengths of specific spectral lines in hydrogen and hydrogen-like ions. The first line of the Lyman series corresponds to the transition from n=2 to n=1 in hydrogen, while the second line of the Balmer series corresponds to the transition from n=4 to n=2 in hydrogen-like ions.

Key Concepts

• Lyman Series: Transitions to the ground state (n=1) from higher energy levels.
• Balmer Series: Transitions to the first excited state (n=2) from higher energy levels.

Wavelength Calculation

The wavelength for these transitions can be calculated using the Rydberg formula:

1/λ = RZ²(1/n₁² - 1/n₂²)

Where R is the Rydberg constant, Z is the atomic number, n₁ is the lower energy level, and n₂ is the higher energy level.

Applying the Formula

For the Lyman series (first line):

n₁ = 1, n₂ = 2, Z = 1 (for hydrogen)

For the Balmer series (second line):

n₁ = 2, n₂ = 4, Z = A (for hydrogen-like ion)

Setting the Wavelengths Equal

Since the wavelengths are equal, we can set the two equations from the Rydberg formula equal to each other and solve for A:

R(1/1² - 1/2²) = R*A²(1/2² - 1/4²)

This simplifies to:

1 - 1/4 = A²(1/4 - 1/16)

3/4 = A²(3/16)

Solving for A gives:

A² = 4, thus A = 2.

Final Answer

The atomic number of the hydrogen-like ion is 2.