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Question

The work function of aluminum is 4.2eV. Light of wavelength 2000˚A is incident on it. The stopping potential for fastest electrons will be

A
Zero
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B
2.025 volt
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C
4.2 volt
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D
6.19 volt
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Solution

The correct option is C 2.025 volt
Energy of incident photons E=hcλ=6.64×1034×3×1082000×1010=9.96×1019J

The energy in electron volts is: E1.618×1019=6.2250eV.

Hence the energy with which the electrons are ejected is :
6.2254.2=2.025eV
That is the stopping potential needed to stop the fastest electron.

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