The maximum velocity of the
photoelectrons emitted from the surface
is v when the light of frequency n falls on a
metal surface. If the incident frequency is
increased to 3n, the maximum velocity of
the ejected photoelectrons will be :
(1) less than 3 v
(2) v
(3) more than 3 v
(4) equal to 3 v
Asked by modi72879
2nd April 2018,
7:16 PM
Answered by Expert
Answer:
Let ν1 be the incident frequency, then we wite the photoelectric equation as hν1 = φ + (1/2)mV2 ....................( 1 )
where V is the velocity of electron that corresponds to maximum kinetic energy, φ is workfunction and h is plancks constant.
if the incident frequency becomes 3ν1, photon energy becomes 3 times of original energy and (1) can be written as
h(3ν1) = φ + ΔE +(1/2)m(√3V)2 ................(2)
In the above equation, the term ΔE should be absorbed in the term representing maximum kinetic energy of electron.
Hence h(3ν1) = φ + (1/2)m (√3V + ΔV)2 ................(3)
Hence from (3), it can be concluded that when frequency is changed to thrice of its original value, maximum velocity of photo-electron will be grater than √3 times of its original value.
( I guess √ is not typed while typing the various options for answer of the question, because we need exact values of φ and ν1 to compare the maximum velocity with 3 times of original maximaum velocity V)
Answered by Expert
3rd April 2018,
1:13 PM
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