If the cathode is cesium - - 1-9 eV- what will be the cut off voltage-

The correct option is D 0.56 VEnergy of photon of wavelength 5000A will be 100005000× 1.23eV=2.46eV because E ∝1λ So maximum kinetic ener ...

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Standard XII

Physics

I vs V - Varying Frequency

If the cathod...

Question

If the cathode is cesium $(ϕ = 1.9 e V),$ what will be the cut off voltage?

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0.46

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0.56

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Solution

The correct option is D $0.56$ V
Energy of photon of wavelength $5000 A$ will be $\frac{10000}{5000} \times 1.23 e V = 2.46 e V$ because $E \propto \frac{1}{λ}$
So maximum kinetic energy of emitted electron will be $E - ϕ = 2.46 - 1.9 = 0.56 e V$
The cutoff potential will be just equal to potential corresponding to maximum kinetic energy of emitted electron
so it will be $\frac{0.56 e V}{e} = 0.56 V$
Option D is correct.

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If the cathode is cesium (ϕ=1.9eV), what will be the cut off voltage?

If the cathode is cesium $(ϕ = 1.9 e V),$ what will be the cut off voltage?