In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm . The decrease in the stopping potential is close to : ( hc e =1240 nm-V)

Correct answer: C. 1.0 ~V Let φ= work function of the metal, hc λ_ 1 =φ+eV_ 1 hc λ_ 2 =φ+eV_ 2 (ii) Sutracting (ii) from (i) we get hc( 1 λ_ 1 - 1 λ_ 2 )=e(V_ 1 -V_ 2 ) V_ 1 -V_ 2 =(h c)/(e)( λ_ 2 -λ_ 1 λ_ 1 · λ_ 2 ) [ array l λ_ 1 =300 nm λ_ 2 =400 nm hc e =1240 nm-V array ] =(1240 nm-v)( 100 nm 300 nm × 400 nm ) =1,03 ~V 1 ~V

JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2019 · 11 January · Evening Shift · Physics · Dual Nature of Matter

In a photoelectric experiment, the wavelength of the light incident...

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from

300 \mathrm{nm}

400 \mathrm{nm}

. The decrease in the stopping potential is close to :

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

Options

A $0.5 \mathrm{~V}$
B $1.5 \mathrm{~V}$
C $1.0 \mathrm{~V}$
D $2.0 \mathrm{~V}$

Correct answer

C. $1.0 \mathrm{~V}$

Step-by-step solution

Let

\phi=

work function of the metal,

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

\frac{\mathrm{hc}}{\lambda_{2}}=\phi+\mathrm{eV}_{2} \quad \ldots \ldots

(ii) Sutracting (ii) from (i) we get

\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)

\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]

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

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

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