The photoelectric threshold wavelength of silver is 3250×10-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×10-10 m is Given h=4.14×10-15eVs and c=3×108 m s-1
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a
≈0.6×106 m s-1
b
≈61×103 m s-1
c
≈0.3×106 m s-1
d
≈6×105 m s-1
answer is A.
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Detailed Solution
Kmax=hv-ϕ0=hv-hvo=hcλ-hcλ0 where λ0= threshold wavelength or 12mv2=hcλ-hcλ0 Here, h=4.14×10-15eVs,c=3×108 m s-1λo=3250×10-10 m=3250A0λ=2536×10-10 m=2536A0,m=9.1×10-31 kghc=4.14×10-15eVs×3×108 m s-1=12420eV\AA∴ 12mv2=1242012536-13250eV=1.076eVv2=2.152eVm=2.152×1.6×10-199.1×10-31∴ v≈6×105 m s-1=0.6×106 m s-1