For intensity I of a light of wavelength 5000 A the photoelectron saturation current is 0-40 - A and stopping potential is 1-36 V- the work function of metal is -

The correct option is B 1.10 eVEnergy due to incoming light is E=hc/λ where h=6.6×10^ 4 is the plank's constant, so #160; E=6.6×10^ 34×3×10^8 ...

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Standard XII

Physics

Dimensional Analysis

For intensity...

Question

For intensity $I$ of a light of wavelength $5000 ˚ A$ the photoelectron saturation current is $0.40 μ A$ and stopping potential is $1.36 V$ , the work function of metal is :

2.47 eV

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1.36 eV

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1.10 eV

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0.43 eV

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Solution

The correct option is B 1.10 eV
Energy due to incoming light is $E = h c / λ$ where $h = 6.6 \times 10^{- 4}$ is the plank's constant, so

$E = 6.6 \times 10^{- 34} \times 3 \times 10^{8} / (5000 \times 10^{- 10} \times 1.6 \times 10^{- 19}) e V = 2.475 e V$

Stopping potential is $1.36 e V$

So work function is $W_{o} = 2.475 - 1.36 = 1.115 e V$

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For intensity I of a light of wavelength 5000˚A the photoelectron saturation current is 0.40μA and stopping potential is 1.36 V, the work function of metal is :

The correct option is B 1.10 eVEnergy due to incoming light is E=hc/λ where h=6.6×10−4 is the plank's constant, so E=6.6×10−34×3×108/(5000×10−10×1.6×10−19)eV=2.475eVStopping potential is 1.36eVSo work function is Wo=2.475−1.36=1.115eV

For intensity $I$ of a light of wavelength $5000 ˚ A$ the photoelectron saturation current is $0.40 μ A$ and stopping potential is $1.36 V$ , the work function of metal is :