Question

A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is ______.

Options

• `8/7lambda`

• `16/7lambda`

• `24/7lambda`

• `32/7lambda`

Answer 1

A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is `underlinebb(32/7lambda)`.

Explanation:

Given: n₁ = 5, n₂ = 1, Z = 1 and λ is wavelength of the wave, R is the Rydberg constant i.e. 1.097 × 10⁷ m^-1 and Z is the atomic number of the element. n_i is lower energy level and n_h is the higher energy level.

Calculating the wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 5 to n = 1 state,

`1/lambda = RZ^2(1/(n_i^2) - 1/(n_h^2))`

⇒ `lambda = RZ^2(1/1^2 - 1/5^2)`

⇒ `lambda = 24/25RZ^2`

⇒ `RZ^2 = (25lambda)/24` ...(i)

Calculating the wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 5 to n = 2 state,

`1/lambda^' = RZ^2(1/n_i^2 - 1/n_h^2)`

`1/lambda^' = RZ^2(1/2^2 - 1/5^2)`

= `RZ^2 xx 21/100`

⇒ `1/lambda^' = (25lambda)/24 xx 21/100` ...[From (i)]

= `32/7 lambda`