Question

For the hydrogen atom emission spectrum, the a line in the Balmer series is produced by the electron transition from n=3 to n=2. The wavenumber of the a line is 15234 cm⁻⁷. What is the wavelength of the Bline produced by the electron transition from n=2 to n=4?

A. 120 nm.
B. 486 nm.
C. 547 nm.
D. 891 nm.

Answer 1

The wavelength of the B line in the Balmer series for the hydrogen atom emission spectrum, produced by the electron transition from n=2 to n=4, is 486 nm.

Step-by-step explanation:

The Balmer series in the hydrogen atom emission spectrum corresponds to electron transitions from higher-energy orbits (n > 2) to the second orbit (n = 2). The wavelength of the a line in the Balmer series is known to be 656 nm. To find the wavelength of the B line produced by the electron transition from n=2 to n=4, we can use the formula:

1/λ = R * (1/n₁² - 1/n₂²)

where λ is the wavelength, R is the Rydberg constant (1.097 x 10^7 m⁻¹), and n₁ and n₂ are the initial and final energy levels, respectively. Substituting the values of n₁=2 and n₂=4 into the formula, we can solve for λ:

1/λ = 1.097 x 10^7 * (1/2² - 1/4²)

Simplifying the equation gives us 1/λ = 1.097 x 10^7 * (1/4 - 1/16).

Calculating this expression yields λ = 486 nm, which corresponds to option B.