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

Question

A and B are two metals with threshold frequencies 1.8× 1014Hz and 2.2× 1014Hz. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Take h=6.6× 10-34J-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× 10-20J =(11.88× 10-20/1.6× 10-19)eV =0.74 text eV Similarly, EB=0.91 text eV Since, 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 2.2×1014Hz. Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by (Takeh=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 Since, 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 $1.8 \times 10^{14} Hz$ and $2.2 \times 10^{14} Hz .$ Two identical photons of energy 0.825 eV each are incident on them. Then photoelectrons are emitted by $(T ak e h = 6.6 \times 10^{- 34} J - s)$