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Q.

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
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