What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum?

Options

A. 102 nm
B. 150 nm
C. 82 nm
D. 122 nm

Answer

D

Explanation

For any series , the transition that produces the least enegetic photon is the transition between the home - base level that defines the series and the level immediately above it . For the Lyman series , the home - base level is at n = 1 . So the transition that produces the least is at n = 1 So te transition that produces the least energetic photon is the transition from the n = 2 level to the n=1 level . ` :. ` The wavelength for the least energetic photon is
` lambda = (hc)/(E_(2)-E_(1))`
Here , `hc = 1240 `eV nm
`E_(1) = -(13.6)/(1^(2))eV = - 13.6 eV `
(as ` E_(n) = - (13.6 )/(n^(2))` eV)
`E_(2) = - (13.6)/(2^(2)) eV = - 3.4eV `
` :. lambda = (1240 "e V nm")/(-3.4 eV -(-13.6 eV))`
` = (1240 " eV" nm )/(10.2 eV) = 122 nm`