The work function of potassium metal is 2.30 eV. When light of frequency 8 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. The stopping potential (in volt) of the electrons will be equal to

Question

The work function of potassium metal is 2.30 eV. When light of frequency 8 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. The stopping potential (in volt) of the electrons will be equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Energy of the incident photon, E=h v E =(6.6 × 10-34 J s )(8 × 1014 Hz ) =52.8 × 10-20 J =(52.8 × 10-20/1.6 × 10-19) eV =3.3 eV According to Einstein's photoelectric equation, e Vs=h v-φ0 e Vs=3.3 eV -2.3 eV =1.0 eV Stopping potential, Vs=1,0 V

Q. The work function of potassium metal is 2.30eV. When light of frequency 8×1014Hz is incident on the metal surface, photoemission of electrons occurs. The stopping potential (in volt) of the electrons will be equal to

Energy of the incident photon, E=hv E=(6.6×10−34Js)(8×1014Hz) =52.8×10−20J=1.6×10−1952.8×10−20​eV=3.3eV According to Einstein's photoelectric equation, eVs​=hv−ϕ0​ eVs​=3.3eV−2.3eV=1.0eV Stopping potential, Vs​=1,0V

Q. The work function of potassium metal is $2.30 e V$ . When light of frequency $8 \times 1 0^{14} Hz$ is incident on the metal surface, photoemission of electrons occurs. The stopping potential (in volt) of the electrons will be equal to