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The emission of electrons from a metal ...

The emission of electrons from a metal surface exposed to light radiation of appropriate wavelength is called photoelectric effect .The emitted electron are called photo-electron work function of threshold energy may be defined as the minimum amount of energy required to eject electrons from a metal surface .According to Einstein
Maximum kinetic energy of ejected electron = Absorbed energy - Work function
`(1)/(2) mv_(max)^(2) = h(v) - h(v_(0)) = hv [(1)/(lambda) - (1)/(lambda_(0))]`
Where `v_(0)` and `lambda_(0)` are threshold frequency and threshold wavelength respectively
Stopping potential : it is the minimum potential at which the photoelectric current becomes zero if `V_(0)` is the stopping potential `eV_(0) = h(v- v_(0))`.
The following figure indicates the energy levels of a certain atom .When the system moves from `2E` level to E level a photon of wavelength `lambda` is emitted .The wavelength of the photon produced during the transition from level `4E//3` to level E is

A

` lambda //3`

B

`4 lambda//3`

C

`3 lambda`

D

`3 lambda//4`

Text Solution

Verified by Experts

The correct Answer is:
C

`lambdav= 2 E -E`
` (hc)/(lambda) = E therefore lambda = (hc)/E`
Also , ` (hc)/(lambda_1) = (4E)/3 - E = E/3`
` :. Lambda_1 = ( hc xx 3)/E = 3 lambda`.
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