A monochromatic radiation of wavelength 975 Å excites the hydrogen atom from its ground state to a higher state
Answers
Here, the question is incomplete . The complete question is that we have to find how many different spectral lines are possible in the resulting spectrum.Which transition corresponds to the longest wavelength ?
Here, it is given that wavelength = 975 Å
= 975 × 10⁻¹⁰ m
As, hydrogen atom is excited from its ground state .
Thus, n₁ = ground state = 1
According to formula -
1/λ = R(1/n₁² - 1/n₂²) , where R = Rydberg constant = 1.09 * 10⁷ m⁻¹
λ = wavelength of radiation emitted
n₂ = excited state
=> 1/975 × 10⁻¹⁰ = R(1/1 - 1/n₂²)
=> 1/975 × 10⁻¹⁰ = 1.09 * 10⁷(1/1 - 1/n₂²)
=> 0.941 = (1/1 - 1/n₂²)
=> n₂ = 4 (approx)
Thus, electron is excited from ground state to 4th energy level .
Now according to the formula -
Number of spectral lines = (n₂ - n₁ + 1)(n₂ - n₁)/2
= (4 - 1 + 1)(4 - 1)/2
= 6
Thus, transition that corresponds to longest wavelength amongst them is
from n = 1 to n = 4 .