A surface irradiated with light of wavelength 480 nm gives out electrons with maximum velocity 2 v m/s, the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity 2 v m/s, if it is irradiated by light of wavelength
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a
300 nm
b
360nm
c
384 nm
d
400 nm
answer is A.
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Detailed Solution
12mv2max=hc480×10−9−hc600×10−9 when 412mv2max=hcλ−hc600×10−9 hcλ−hc600×10−9=4hc480×10−9−hc600×10−9 Solving, we get λ=300nm.