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Question

Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength 2000A falls on the surface, the velocity of the fastest electron ejected from the surface will be


A

8.4×105 m/sec

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B

7.4×105 m/sec

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C

6.4×105 m/sec

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D

8.4×105 m/sec

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Solution

The correct option is A

8.4×105 m/sec


By using E=W0+12mv2max where E=123752000=6.18eV
6.18eV=4.2eV+12mv2max1.98eV=12mv2max
1.98×1.6×1019=12×9.1×1031×v2max
vmax=8.4×105 m/s


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