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Question

Electromagnetic radiation having λ=310 ˙A is subjected to a metal sheet having work function = 12.8 eV. What will be the velocity of photo electrons having maximum kinetic energy?

A
1.23×106m/s
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B
2.28×106m/s
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C
3.1×106m/s
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D
None of these
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Solution

The correct option is C 3.1×106m/s

Energy of photon of wavelength, λ=310×1010 is hcλ.

Therefore ,E(photon)=(6.62×1034)(3×108)(31×109)=6.406×1018J.

Work function, W=12.8eV=12.6×1.6x1019J=2.016×1018J.

Now, maximum kinetic energy of photo electron,

(0.5)mv2(max)=E(photon)W=6.406×10182.016×1018J=4.39×1018J.

Therefore,v2(max)=(4.39×1018)(2)(9.1×1031)=9.648×1012m2s2.

Therefore, v(max)=3.106×106ms1

Hence,option C is correct answer.


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