The difference between the incident energy and threshold energy for an electron in a photoelectric effect experiment is 5eV. The de Broglie wavelength of the electron is -
A
6.6×10−9√1456 m
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B
6.6×10−9√145.6 m
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C
6.6×10−9√1664
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D
6.6×10−9√166.4m
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Solution
The correct option is B6.6×10−9√145.6 m K.E of e− ejected = Energy of incident quantum - threshold energy = 5 eV