A beam of momechromatic light of wavelength `lambda` ejectes photonelectrons from a cesium `(phi = 1.9eV)` these photonelectron are mde to collide with h-atom in ground state . find the maximum value of `lambda` for which (a) hydrogen atoms may be ionised (b) hydrogen may get excited from the ground state to the first excited state and (c ) the excited hydrogen atoms may emit visible light

Answer

A, B, C

Explanation

Here `phi = 1.9 eV`
(a) The hydrogen is ionized
`n_1 = 1`, `n_2=oo`
Energy required for ionization
`=13.6((1)/(n_1^2)-(1)/(n_2^2))=13.6`
:. `(hc)/lambda-1.9 =13.6`
implies ` 1242/lambda = 15.5`
implies `lambda = 1242/15.5`
`=80.1 nm = 80 nm`
For the electron to be excited from
`n_1 = 1` to `n_2=2`
`E=13.6((1)/(n_1^2)-(1)/(n_2^2))`
`=13.6(1-(1)/(4))`
`=(13.66 xx 3)/(4)`
:. `(hc)/(lambda)-1.9 = (13.6 xx3)/(4)`
implies `(hc)/lambda =(13.6 xx 3)/(4) + 1.9`
implies `1242/lambda = 10.2 +1.9 =12.1`
implies `lambda = 1242/12.1`
`=102.64 = 102 nm`