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Question

When a certain metal was irradiated with light of frequency 3.2×1016 Hz, the photoelectrons emitted had twice the kinetic energy as that of the photoelectrons emitted when the same metal was irradiated with light of frequency 2.0×1016 Hz. The value of ν for the metal was found to be x×1016 Hz.
Then x=

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Solution

According to photoelectric effect:
hν = w + K.E.
where,
h = Planck's constant
hν = Energy of photon
w = Work function = hνo
νo = threshold frequency
K.E. = Kinetic energy of electrons released from the metal surface

As per the given statements,
h ( 3.2 ×1016)=w+2K.E ........(1)
h ( 2×1016)=w+K.E. ............(2)

Subtracting (2) from (1) we get,
h (1.2 ×1016) =K.E.
6.626×1034×1.2×1016=7.95×1018=K.E.
Put value of K.E in equation (2)
h ( 2×1016)=hνo+K.E.
(6.626×1034×2×1016)=hν +7.95×1018
13.25×10187.95×10186.626×1034 = ν
On solving, νo = 7.99×1015 = 0.8×1016 Hz

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