Photons of energy 6 eV are incident on a potassium surface of work function 2.1 eV. The stopping potential will be

Question

Photons of energy 6 eV are incident on a potassium surface of work function 2.1 eV. The stopping potential will be (A) -8.1 V (B) -3.9 V (C) -6 V (D) -2.1 V

Tardigrade Admin · Accepted Answer

Ek=h v-φ=6-21=3.9 eV Stopping potential (Ek/e)=-(3.9 eV / e ) =-3.9 V

Q. Photons of energy $6 e V$ are incident on a potassium surface of work function $2.1 e V$ . The stopping potential will be

-8.1 V

-3.9 V

-6 V

-2.1 V

$E_{k} = h v - ϕ = 6 - 21 = 3.9 e V$
Stopping potential $\frac{E _{k}}{e} = - \frac{3.9 e V}{e}$
$= - 3.9 V$

Q. Photons of energy 6eV are incident on a potassium surface of work function 2.1eV. The stopping potential will be