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The stopping potential for photoelectrons emitted from a surface illuminated by light wavelength of 5893 Å is 0.36 V. Calculate the maximum kinetic energy of photoelectrons.

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1.746 eV

1.386 eV

2.106 eV

0.36 eV

answer is D.

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Detailed Solution

We know that

${KE}_{\max} = hf - ϕ = (\frac{hc}{λ}) - ϕ$

$or ϕ = \frac{hc}{λ} - {KE}_{max}$

$Also, {KE}_{max} = {eV}_{s} = 0 .36 eV$

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