A and B are two metals with threshold frequencies 1.8 × 1014 Hz and 2.2 × 104 Hz Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Take .h=6.6 × 10-34 J - s )

Question

A and B are two metals with threshold frequencies 1.8 × 1014 Hz and 2.2 × 104 Hz Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Take .h=6.6 × 10-34 J - s ) (A) B alone (B) A alone (C) Neither A nor B (D) Both A and B

Tardigrade Admin · Accepted Answer

Threshold energy of A is EA=hvA =6.6 × 10-34 × 1.8 × 1014 =11.88 × 10-20 J =(11.88 × 10-20/1.6 × 10-19) eV =0.74 e V Similarly, EB=0.91 eV As the incident photons have energy greater than EA but less than EB. So, photoelectrons will be emitted from metal A only

Q. A and B are two metals with threshold frequencies 1.8×1014Hz and 2.2×104Hz Two identical photons of energy 0.825eV each are incident on them. Then photoelectrons are emitted by (Take h=6.6×10−34J−s)

B alone

A alone

Neither A nor B

Both A and B

Threshold energy of A is EA​=hvA​ =6.6×10−34×1.8×1014 =11.88×10−20J =1.6×10−1911.88×10−20​eV =0.74eV Similarly, EB​=0.91eV As the incident photons have energy greater than EA​ but less than EB​. So, photoelectrons will be emitted from metal A only

Q. $A$ and $B$ are two metals with threshold frequencies $1.8 \times 1 0^{14} Hz$ and $2.2 \times 1 0^{4} Hz$ Two identical photons of energy $0.825 e V$ each are incident on them. Then photoelectrons are emitted by (Take $h = 6.6 \times 1 0^{- 34} J - s)$