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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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