The wavelength of the radiation emitted is when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be . The value of is __________.
Show answer and solution
Answer: 27
Solution
The second excited state is n=3, first excited state is n=2. Energy difference: E∝(1/4 - 1/9) = 5/36, so hc/λ_o ∝ 5/36.
The third excited state is n=4, second orbit is n=2. Energy difference: ∝(1/4 - 1/16) = 3/16.
The ratio of wavelengths: λ/λ_o = (5/36)/(3/16) = (5×16)/(36×3) = 80/108 = 20/27.
So λ = (20/27)λ_o, meaning x = 27.
The third excited state is n=4, second orbit is n=2. Energy difference: ∝(1/4 - 1/16) = 3/16.
The ratio of wavelengths: λ/λ_o = (5/36)/(3/16) = (5×16)/(36×3) = 80/108 = 20/27.
So λ = (20/27)λ_o, meaning x = 27.
[JEE Main 2023 · 25 Jan · Shift 1]