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प्रश्न
If the frequency of the incident radiation is increased from 4 × 1015 Hz to 8 × 1015 Hz, by how much will the stopping potential for a given photosensitive surface go up?
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उत्तर
The above equation can be written as
hv - hv0 = eVs
hv1 - hv0 = eV1 ....(i)
hv2 - hv0 = eV2 ....(ii)
Subtracting (i) and (ii)
h( v2 - v1 ) = e( V2 - V1 )
∴ Change in stopping potential = `(h( v_2 - v_1 ))/e`
= `( 6.6( 8 xx 10^15 - 4 xx 10^15 ))/( 1.6 xx 10^-19 ) xx 10^-34`
= `( 6.6 xx 4 xx 10^15 )/( 1.6 xx 10^-19 ) xx 10^-34`
= `( 6.6 xx 4 )/1.6`
= 16.5 V
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