In the arrangement shown in Fig, $y = 1.0 m m, d = 0.24 m m$ and $D = 1.2 m$ . The work function of the material of the emitter is $2.2 e V$ . Find the stopping potential $V$ needed to stop the photocurrent. If the answer is $x \times 10^{- 1} V$ , report x.

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Solution

The distance between consecutive bright and dark fringes is given by : $y$ = $\frac{λ D}{2 d}$
$λ$ = $\frac{2 d y}{D}$ = $\frac{2 \times 0.24 \times 1 \times 10^{- 6}}{1.2}$ = $4 \times 10^{- 7} m = 400 n m$
$e V_{s} = \frac{h C}{λ} - ϕ = \frac{1240}{400} - 2.2 = 0.9 e V$
$∴ V_{s} = 0.9 = 9 \times 10^{- 1}$

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In the arrangement shown in Fig, y=1.0mm,d=0.24mm and D=1.2m. The work function of the material of the emitter is 2.2eV. Find the stopping potential V needed to stop the photocurrent. If the answer is x×10−1V, report x.

The distance between consecutive bright and dark fringes is given by : y = λD2dλ = 2dyD = 2×0.24×1×10−61.2 = 4×10−7m=400nmeVs=hCλ−ϕ=1240400−2.2=0.9eV∴Vs=0.9=9×10−1

In the arrangement shown in Fig, $y = 1.0 m m, d = 0.24 m m$ and $D = 1.2 m$ . The work function of the material of the emitter is $2.2 e V$ . Find the stopping potential $V$ needed to stop the photocurrent. If the answer is $x \times 10^{- 1} V$ , report x.