Question

($h=6.63×10_{−34}$Js and $c=3×10_{8}$ m/s)

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Solution

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Correct option is B)

Given:

$λ=2500×10_{−10}m$

$E=2500×10_{−10}6.63×3×10_{−26} J$

$E=1.6×10_{−19}0.007956×10_{−16} eV=4.97eV$

Work function of metal is $3.5eV$.

So maximum kinetic energy of the photoelectron after emission $K=E−W=(4.97−3.5)eV=1.47eV$

So the voltage required to stop the fastest photoelectron is $1.47V$.

So the voltage required to stop the fastest photoelectron is $1.47V$.

So, the answer is option (B).

Video Explanation

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