In the arrangement shown in figure, y =1.0 mm , d =0.24 mm and D =1.2 m. The work function of the material of the emitter is 2.18 eV. The stopping potential needed to stop the photocurrent is volts. (Take h =4.14 × 10-15 eVs )

Question

In the arrangement shown in figure, y =1.0 mm , d =0.24 mm and D =1.2 m. The work function of the material of the emitter is 2.18 eV. The stopping potential needed to stop the photocurrent is volts. (Take h =4.14 × 10-15 eVs ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The distance between bright band and dark band is given by,

y = \frac{λ D}{2 d}

∴ λ = \frac{2 y d}{D}

= \frac{2 \times 1 0 ^{- 3} \times 0.24 \times 1 0 ^{- 3}}{1.2}

= 4 \times 1 0^{- 7} m

E = \frac{h c}{λ}

= \frac{4.14 \times 1 0 ^{- 15} \times 3 \times 1 0 ^{8}}{4 \times 1 0 ^{- 7}}

= 3.105 e V

Stopping potential,

e V_{0} = 3.105 - 2.18 = 0.925 e V

V_{0} = 0.925 V \approx 0.92 V

Q. In the arrangement shown in figure, y=1.0mm,d=0.24mm and D=1.2m. The work function of the material of the emitter is 2.18eV. The stopping potential needed to stop the photocurrent is ________volts. (Take h=4.14×10−15eVs )

The distance between bright band and dark band is given by, y=2dλD​ ∴λ=D2yd​ =1.22×10−3×0.24×10−3​ =4×10−7m E=λhc​ =4×10−74.14×10−15×3×108​ =3.105eV Stopping potential, eV0​=3.105−2.18=0.925eV V0​=0.925V≈0.92V

Q. In the arrangement shown in figure, $y = 1.0 mm, d = 0.24 mm$ and $D = 1.2 m$ . The work function of the material of the emitter is $2.18 e V$ . The stopping potential needed to stop the photocurrent is ________volts. (Take $h = 4.14 \times 1 0^{- 15} e V s$ )