Energy of first Balmer line of hydrogen atom is x J. Find energy of second Balmer line.

Solution:

For Balmer series:

First Balmer line: n = 3 → n = 2

Second Balmer line: n = 4 → n = 2

Energy emitted:

E = 13.6 × (1/n₁² - 1/n₂²) eV

First Balmer line:

E₁ = 13.6 × (1/2² - 1/3²)

E₁ = 13.6 × (5/36)

Second Balmer line:

E₂ = 13.6 × (1/2² - 1/4²)

E₂ = 13.6 × (3/16)

Now,

E₂ / E₁ = (3/16) ÷ (5/36)

E₂ / E₁ = 27/20

E₂ = (27x)/20

Answer: (27x/20) J