Explanation
The frequency of incident light is less than the threshold frequency. No photoelectrons will be emitted from the metal surface. The threshold frequency is 7.24×1014Hz, and no photoelectron will be ejected.
The energy of a neutron in eV whose de-Broglie wavelength is 10A
A non-monochromatic light is used in an experiment on the photoelectric effect. The stopping potential is related to the:
Powerofabulb(p)=40Wenergyemittedbyabulb=power×Time(s)=40×20=800JEnergyofphotonsemittedbyabulb=(80100)×800=640JWavelength(λ)=620nmEnergyofaphoton=hcλ=12400×1.6×10−19×10−10620×10−9=6403.2×10−19=2×1021photons
Hence,
option (D) is correct answer.
λ=hpandp22m=ε=32kT∴p=(3mkT)12λ=hp=h(3kmT)12∴T=h23kmλ2T=h23kλ2=(6.626×10−34)2(3)(1.38×10−23)(9.108×10−31)(76.3×10−9)2=2.00KT=2.00K
Option B is correct .
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