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Question

The photoelectric thresshold 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 approximately
(Given h=4.14×1015 eV and c=3×108 ms1)


A
0.3×106ms1
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B
6×105ms1
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C
6×106ms1
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D
61×103ms1
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Solution

The correct option is B 6×105ms1

Kinetic Energy, 12mv2=hc[1λ1λ0]
Also we can write in terms of frequency,
12mv2=hvhvo

hv=4.14×1015×3×1032536×1010=4.897 eV

hv0=4.14×1015×3×1033250×1010=3.32 eV

12mv2=4.8973.32=1.077 eV

v=2×1.077×1.6×10199.1×1031

v=0.615×106 m/s


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