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Question

When a certain metal was irradiated using a light of frequency 8.1×1016 Hz the photoelectron emitted had 1.5 times the kinetic energy as did the photoelectrons emitted when the same metal was irradiated using a light of frequency 5.8×1016 Hz. If the same metal is irradiated using light of wavelength 3.846 nm, what will be the kinetic energy of the photoelectron emitted?


A
180 eV
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B
3.65×1017 J
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C
2.28×102 eV
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D
4.37×1017 J
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Solution

The correct option is D 4.37×1017 J
Given Information:
Frequency ν1=5.8×1016 Hz
Kinetic energy = K.E1
Frequency ν2=8.1×1016 Hz
Kinetic energy = 1.5×K.E1
work function = hνK.E
so we get,
Work function = h×5.8×1016K.E1
Work function = h×8.1×10161.5×K.E1
On solving,
Work function =7.944×1018 J
Kinetic energy =3.045×1017 J
Wavelength of the incident radiation =38.46 A
Energy of the incident radiation = hν
=6.626×1034×3×10838.46×1010
=5.16×1017 J

Kinetic energy, =5.16×10177.94×1018 J
=4.37×1017 J

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