In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300nm to 400nm. The decrease in the stopping potential is close to: (hce=1240nm−V)
A
0.5V
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B
1.0V
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C
2.0V
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D
1.5V
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Solution
The correct option is A1.0V hcλ1=ϕ+eV1 .(i) hcλ2=ϕ+eV2 (ii) (i)-(ii) hc(1λ1−1λ2)=e(V1−V2) ⇒V1−V2=hce(λ2−λ1λ1−λ2) =(1240nm−V)100nm300nm×400nm =1V.