Photoelectric EffectHard
Question
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speed of emitted electrons will be
Options
A.1 : 4
B.1 : 2
C.1 : 1
D.1 : 5
Solution
Here, work function, φ0 = 0.5 eV
According to Einstein′s photoelectric equation
Maximum kinetic energy = Incident photon - Work function
of emitted electron energy
∴ Kmax1 = 1 eV - 0.5 eV = 0.5 eV ........(i)
and Kmax2 = 2.5 eV - 0.5 eV = 2.0 eV ........(ii)
Divide (i) by (ii), we get
According to Einstein′s photoelectric equation
Maximum kinetic energy = Incident photon - Work function
of emitted electron energy
∴ Kmax1 = 1 eV - 0.5 eV = 0.5 eV ........(i)
and Kmax2 = 2.5 eV - 0.5 eV = 2.0 eV ........(ii)
Divide (i) by (ii), we get
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