In a photoemissive cell, with exciting wavelength `lambda`, the faster electron has speed v. If the exciting wavelength is changed to `3lambda//4`, the speed of the fastest electron will be
A
Less then ` v(4//3)^(1//2)`
B
`v(4//3)^(1//2)`
C
`v(3//4)^(1//2)`
D
Greater then ` v(4//3)^(1//2)`
Text Solution
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The correct Answer is:
D
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