Q.
The photoelectric threshold wavelength for a certain metal is4000A°. The maximum energy of the ejected photoelectrons by a radiation of 2000A° is
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a
2.13 eV
b
3.1 eV
c
1.23 eV
d
3.82 eV
answer is B.
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Detailed Solution
hcλ=hcλ0+kE kE=12400(12000−14000) =124004000=3.1ev
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