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Question

Physics Question on Photoelectric Effect

AA and BB are two metals with threshold frequencies 1.8×1014Hz1.8 \times 10^{14}\, Hz and 2.2×104Hz2.2 \times 10^{4}\, Hz Two identical photons of energy 0.825eV0.825 \,eV each are incident on them. Then photoelectrons are emitted by (Take h=6.6×1034Js)\left.h=6.6 \times 10^{-34} \,J - s \right)

A

B alone

B

A alone

C

Neither A nor B

D

Both A and B

Answer

A alone

Explanation

Solution

Threshold energy of AA is
EA=hvAE_A=hv_A
=6.6×1034×1.8×1014=6.6 \times 10^{-34} \times 1.8 \times 10^{14}
=11.88×1020J=11.88 \times 10^{-20} \,J
=11.88×10201.6×1019eV=\frac{11.88 \times 10^{-20}}{1.6 \times 10^{-19}} \,eV
=0.74eV=0.74 \,e V
Similarly, EB=0.91eVE_{B}=0.91 \,eV
As the incident photons have energy greater than EAE_{A} but less than EBE_{B}.
So, photoelectrons will be emitted from metal AA only