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प्रश्न
Obtain an expression for de-Broglie wavelength of wave associated with material particles. The photoelectric work function for metal is 4.2 eV. Find the threshold wavelength.
उत्तर
De-Broglie equated the energy equation of Plank (wave nature) and Einstein (particle nature) such that,
E = hv (Plank energy relation)
E = mc2 (Einsteins mass-energy relation)
Where,
E = energy associated with the particle
h = planks constant
v = frequency associated with the particle
m = moss of the particle
c = speed of light
After equating equations (1) and (2) we get:
hv = mc2
`"h" "c"/lambda = "mc"^2 ...(because "v" = "c"/lambda)`
`lambda = "hc"/"mc"^2 = "h"/"mc"`
If the particle is moving with velocity 'v' then equation (3) becomes,
`lambda = "h"/"mv"`
The wavelength of the particle if energy is in electron volt is,
`lambda = (12,400 Å)/"E"`
Where E should be in eV.
After substituting the value of E i.e. 4.2 eV in equation (4) we get:
`lambda = (12,400 Å)/4.2`
= 2952.38 Å
Hence, the threshold wavelength of the particle is 2952.38 Å.
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