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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`
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Solution
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. ni is lower energy level and nh 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`
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