Question

Draw a neat and labelled energy level diagram and explain Balmer series and Bracket series of spectral lines for hydrogen atom.
The work function for a metal suface is 2.2 eV. If light of wavelength 5000 Å is incident on the surface of metal, find the threshold frequency and incident frequency. Will there be an emmission of photoelectrons or not ?
`(c=3xx10^(8)m//s, 1eV=1.6xx10^(-19)J, h=6.63 xx10^(-34)J.s.)`

Answer 1

Energy level diagram:

Baelmer series :
(i) spectral lines of this series correspond to the transition of an electron from some higher energy state to `2^(nd)` orbit.
(ii) For Balmer series, p = 2 and n = 3, 4, 5, …….
The wave numbers and the wavelengths of spectral lines constituting the Balmer series are given by,
`bar x=(1)/(lambda)=R((1)/(2^(2))-(1)/(n^(2)))`
(iii) There are infinite number of lines in this series out of which four lines are seen called `H_(alpha),H_(beta), H_(gamma), H_(delta).`
(iv) The series lies in the visible region. Wavelengths for n = 3 and are 6563 Å and 4868 Å respectively.
Brackett series :
(i) The spectral lines of this series corresponds to the transition of an electron from a higher energy state to the `4^(th)` orbit.
(ii) For this series, p = 4 and n = 5, 6, 7, .....
The wave numbers and the wavelengths of the spectral lines constituting the Brackett series are given by,
`bar v=(1)/(lambda)=R((1)/(4^(2))-(1)/(n^(2)))`
(iii) This series lie in the near infrared region of the spectrum and contains infinite number of lines.
Wavelengths for n = 5 and 6, are 40518 Å and 26253 Å respectively.
Numerical : Given,
`phi_(0)=2.2 eV`
`=2.2xx1.6 xx 10^(-19)`
`=3.52 xx 10^(-19)J`
`lambda=5000 Å `
`=5xx10^(-7)m`
`h=6.63 xx 10^(-34)J-s`
Frequency of incident radiation,
`v=(c)/(lambda)=(3xx10^(8))/(5xx10^(-7)) `
`=6xx10^(14)Hz`
And, work function for a metal
`phi_(0)=hv_(0)`
or `v_(0)=(phi_(0))/(lambda)`
`=(3.52xx10^(-19))/(6.63xx10^(-34))`
`=5.31 xx 10^(14)Hz`
since, `v gt v_(0)`
`therefore` Emission of photoelectrons will take place.