Dual Nature of Matter

JEE Main level, single correct, medium

175 of 200

Question 175 of 200

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300nm300 \mathrm{nm} to 400nm400 \mathrm{nm}. The decrease in the stopping potential is close to :

(hce=1240nm−V)\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\right)

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Answer: (C) 1.0 V1.0 \mathrm{~V}

Let ϕ=\phi= work function of the metal,

hcλ1=ϕ+eV1\frac{\mathrm{hc}}{\lambda_{1}}=\phi+\mathrm{eV}_{1}

hcλ2=ϕ+eV2……\frac{\mathrm{hc}}{\lambda_{2}}=\phi+\mathrm{eV}_{2} \quad \ldots \ldots

(ii)

Sutracting (ii) from (i) we get

hc(1λ1−1λ2)=e(V1−V2)\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)

⇒V1−V2=hce(λ2−λ1λ1⋅λ2)[λ1=300nmλ2=400nmhce=1240nm−V]\Rightarrow V_{1}-V_{2}=\frac{h c}{e}\left(\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1} \cdot \lambda_{2}}\right)\left[\begin{array}{l}\lambda_{1}=300 \mathrm{nm} \\ \lambda_{2}=400 \mathrm{nm} \\ \frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\end{array}\right]

=(1240nm−v)(100nm300nm×400nm)=(1240 \mathrm{nm}-\mathrm{v})\left(\frac{100 \mathrm{nm}}{300 \mathrm{nm} \times 400 \mathrm{nm}}\right)

=1,03 V≈1 V=1,03 \mathrm{~V} \approx 1 \mathrm{~V}

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