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Q.

A graph is drawn between stopping potential (y-axis) and incident frequency (x-axis). The intercept on x-axis gives 0.5 × 1015 Hz. The K.E. of photo electron emitted from the metal, when the energy of incident photon is 5 eV is

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a

7 eV 

b

3 eV

c

1.5 eV   

d

6 eV

answer is B.

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Detailed Solution

Intercept on x-axis gives threshold frequency νo

⇒KEmax=E−W0=E−hvo

=5 eV−6.6×10−34×0.5×10151.6×10−19eV≃3 eV

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