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Question

The photoelectric threshold wavelength of silver is 3250×1010 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×1010 m is

A
0.6×106 m/s
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B
61×103 m/s
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C
0.3×106 m/s
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D
8×108 m/s
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Solution

The correct option is A 0.6×106 m/s
The maximum kinetic energy is given as,
Kmax=hvϕ0=hvhv0=hcλhcλ0where λ0=threshold wavelengthor 12mv2=hcλhcλ0
Here,
h=4.14×1015 eV,c=3×108m/sλ0=3250×1010 m=32500Aλ=2536×1010 m=25360A,m=9.1×1031Kg
So,
hc=(4.14×1015 ) × (3×108) =12420 eV 0A12mv2=12420(1253613250]eV = 1.076 eVv2=2.152 eVm=2.152×1.6×10199.1×1031v 0.6×106 m/s

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