Question

# In a photo-emissive cell, with exciting wavelength λ, the fastest electrons has speed μ. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be

A
υ(3/4)1/2
B
υ(4/3)1/2
C
Less than υ(4/3)1/2
D
Greater than υ(4/3)1/2
Solution
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#### We know that,hν−ωo=12mv2max=>hcλ=hcλo=12mv2max=> hc(λo−λλλo)=12mv2maxvmax=√2hcm(λo−λλλo)---------- (1)When wavelength is 3λ4 and velocity is v'then v/=√2hcm[λo−(3λ/4)(3λ/4)×λo]−−−−−−−(2)Dividing equation, (2) by (1),v/v=√[λo−(3λ/4)](34)λλo×λλoλo−λv/=v(43)12√[λo−(3λ/4)]λo−λthat is, v/=v(43)12

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