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Question

Radiation of wavelength λ is incident on a photocell. The fastest emitted electron has a speed v. If the wavelength is changed to 3λ4, the speed of the fastest emitted electron will be :
  1. v>(43)1/2
  2. v<(43)1/2
  3. v=(43)1/2
  4. v=(34)1/2

A
v>(43)1/2
B
v<(43)1/2
C
v=(43)1/2
D
v=(34)1/2
Solution
Verified by Toppr

We know,

hcλϕ=12mv2(i)

4hc3λϕ=12mv2(ii)

(ii) - (i) gives

hc3λ=12m(v2v1)

v1=v2+2hc3λm(iii)

also from (i) hcλ=ϕ+mv22

2hcλm=2ϕmv2

2hc3λm=2ϕ3m+r23>v23 (iv)

v>4v23

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