Q.
Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength 2000 Ao falls on the surface, the velocity of the fastest electron ejected from the surface will be
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a
8.4×105m/sec
b
7.4×105m/sec
c
6.4×105m/sec
d
8.4×106m/sec
answer is A.
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Detailed Solution
By using E=Wo+12mvmax2whereE=123752000=6.18eV⇒6.18eV=4.2eV+12mvmax2⇒1.98eV=12mvmax2⇒1.98×1.6×10−19=12×9.1×10−31×vmax2⇒vmax=8.4×105m/s
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