λ1=350nm,λ2=540nm,V2V1=2⇒V1=2VandV2=V
λ1hc=ϕ+21m(2V)2
λ2hc=ϕ+21mV2
λ1hc−4λ2hc=ϕ−4ϕ
hc(5404−3401)=3ϕ
ϕ=1.8eV
JEE Main 2019 — Physics Modern Physics
The surface of certain metal is first illuminated with light of wavelength λ1=350nm and then, by a light of wavelength λ2=540nm. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to
(Energy of photon =λ(innm)1240eV )
Held on 9 Jan 2019 · Verified 6 Jul 2026.
2.5
1.8
5.6
1.4
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