In a photo-missive cell, with exciting wavelength λ the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4 the speed of the fastest emitted electron will be
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a
V(3/4)1/2
b
v(4/3)1/2
c
less than v(4/3)1/2
d
greater than v(4/3)1/2
answer is D.
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Detailed Solution
hcλ1=W0+12mv12 and hcλ2=W0+12mv22 These expressions show that v2∝1/λ v1v2=1/λ11/λ2=λ2λ1 =3λ/4λ=341/2 or v2=v1431/2