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Q.

Ultraviolet light of 6.2 eV falls on an aluminium surface (work function = 4.2 eV). The kinetic energy (in joule) of the fastest electron emitted is approximately

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a

3  ×  10−21

b

3  ×  10−19

c

3  ×  10−17

d

3  ×  10−15

answer is B.

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Detailed Solution

hv=hv0  +  KEKE=hv−hv0    hv=6.2  eV , W=    hv0=4.2  eVK.E=6.2−4.2 =2eV=2.00 × 1.6 ×  10−19KE=3.2  ×  10−19 J
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