Draw a neat, labelled energy level diagram for H atom showing the transitions.
Explain the series of spectral lines for H atom. Whose fixed inner orbit numbers are 3 and 4 respectively.
The work function for potassium and caesiumare 2.25 eV and 2.14 eV respectively. Is the photoelectric effect possible for either of them if incident wavelength is 5180 Å ?
[Given : Planck's constant `=6.63 xx 10^(-34) J.s.`, Velocity of light `=3xx10^(8)m//s, 1 eV =1.6xx10^(-19)J`]

Explanation

Diagram: Refer to HSC Paper, February 2018, Answer 8 (OR).
The series of spectral lines for H atom, whose fixed inner orbit numbers are 3 and 4 Paschen and Brackett series respectively.
Paschen Series : This series originates due to transitions of the electrons from different outer orbits to the `3^(rd)` orbit (p = 3). Therefore, the wavelength of Paschen series is given by
`(1)/(lambda)=R((1)/(3^(2))-(1)/(n^(2)))`
where, n = 4, 5, 6, 7 ...................
This series lies in infrared region of the spectrum.
Brackett Series : This series originates due to transitions of the electrons from different outer orbits to the `4^(th)` orbit (p = 4). Therefore, the wavelength of Brackett series is given by
`(1)/(lambda)=R((1)/(4^(2))-(1)/(n^(2)))`
where, n = 5, 6, 7, 8, ........
This series lies in near - infrared region of the spectrum.
Numerical :
Given : work function for potassium
`(phi_(0))=2.25 eV`
`=2.25 xx 1.6xx10^(-19)J`
`=3.6 xx 10^(-19)J`
and work function for caesium `(phi_(0)^('))`
`=2.14eV`
`=2.14xx1.6xx10^(19)J`
`=3.424 xx 10^(-19)J`
`lambda =5180` Å
`=5.180 xx 10^(-7)m`
Now, `v=(c)/(lambda)`
`=(3xx10^(8))/(5.180xx10^(-7))`
`=0.579xx10^(15)Hz`
`v_(0)=(phi_(0))/(h)`
`=(3.6xx10^(-19))/(6.63xx10^(-34))`
`=0.543 xx 10^(15)Hz`
`and v_(0)^(')=(phi_(0)^('))/(h)`
`=(3.424xx10^(-19))/(6.63xx10^(-34))`
`=0.516xx10^(15)Hz`
As, `v gt v_(0)` and also `v gt v_(0)^(') `
Hence, photoelectric effect is possible for both potassium as well as caesium.