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Question:

Radiation of wavelength λ is incident on a photocell. The fastest emitted electron has speed υ. If the wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be:

=υ(4/3)1/2

<υ(4/3)1/2

=υ(3/4)1/2

υ(4/3)1/2

Solution:

Kinetic energy of emitted electron
E1=hc/λ - φ
where λ is incident wavelength of radiation and φ is the work function of the photocell.
∴(1/2)mv2 = hc/λ - φ. (i)
In second case : Incident wavelength is 3λ/4.
Thus we get
E2 = 4hc/(3λ) - φ
Or (1/2)mv22 = 4hc/(3λ) - φ (ii)
Clearly, E2 > (4/3)E1 and hence v2 > v(4/3)1/2.