Q.

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

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a

3×1021

b

3×1019

c

3×1017

d

3×1012

answer is B.

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Detailed Solution

hv=W0+K.E 62=42+KE.  or  KE.=2eV or K.E.=2×16×1034J=32×1019J

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