In a photo-emissive cell, with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ4,the speed of the fastest emitted electron will be
A
less than v(4/3)1/2
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B
v(4/3)1/2
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C
v(3/4)1/2
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D
greater than v(4/3)1/2
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Solution
The correct option is D greater than v(4/3)1/2 According to Einstein's photoelectric equation, 12mv2=hcλ−W0and12mv12=hc3λ/4−W0=43(hcλ−34W0)=43(hcλ−W0+14W0)=43(12mv2+14W0)(] So, v1 is greater than v(4/3)1/2.