Question

# In a photoemissive cell with exciting wavelength 位 , the fastest electron has a speed v. If the exciting wavelength is changed to 3位4, the speed of the fastest emitted electrons will be

A
V(34)12
B
V(43)12
C
<V(43)12
D
>V(43)12
Solution
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#### K.E.max= incident energy - work function12mv2=hcλ−ϕNow, wavelength is changed to 3λ4so, incident energy =hc3λ/4=4hc3λso, K.E.max=43hcλ−ϕK.E.max=43hcλ−ϕ−ϕ3+ϕ312mv21=43(hcλ)−ϕ+ϕ3=43(12)mv2+ϕ3so,v21>43v2or v1>√43v

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