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Question
A beam of monochromatic light of wavelength λ ejects photoelectrons from a cesium surface (Φ = 1.9 eV). These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of λ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.
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Solution
Given:
Work function of cesium surface, ϕ = 1.9 eV
(a) Energy required to ionise a hydrogen atom in its ground state, E = 13.6 eV
From the Einstein's photoelectric equation,
`(hc)/lamda = E + Ø`
Here,
h = Planck's constant
c = Speed of light
λ = Wavelength of light
`therefore (hc)/lamda = 1.9 = 13.6`
`rArr 1240/lamda = 15.5`
`rArr lamda =(1240)/15.5 `
`rArr lamda = 80 nm`
(b) When the electron is excited from the states n1 = 1 to n2 = 2, energy absorbed (E1) is given by
`E_1 = 13.6 (1/n_1^2 - 1/n_2^2)`
`E_1 = 13.6 (1- 1/4)`
`E_1 = (13.66xx3)/4`
For Einstein's photoelectric equation,
`therefore (hc)/lamda - 1.9 = (13.6xx3)/4`
`rArr (hc)/lamda = (13.6xx3)/4 + 1.9`
`1240/(lamda) = 10.2 + 1.9 =12.1`
`rArr lamda = (1240)/12.1`
`lamda = 102.47 = 102 nm `
(c) Excited atom will emit visible light if an electron jumps from the second orbit to third orbit, i.e. from n1 = 2 to n2 = 3. This is because Balmer series lies in the visible region.
Energy (E2) of this transition is given by
`E_2 = 13.6 (1/n_1^2 - 1/n_2^2)`
`E_2 = 13.6 (1/4 - 1/9)`
`E_2 = (13.66xx5)/(36)`
For Einstein's photoelectric equation,
`(hc)/lamda - 1.9 = (13.6xx5)/36`
`rArr (hc)/lamda = (13.6xx5)/36+ 1.9 `
`rArr 1240/lamda = 1.88 + 1..9 = 3.78`
`rArr lamda = 1240/3.78`
`rArr lamda = 328.04 nm
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