Question

The work function of $Pt$ is twice as the work function of $Ca$. If the minimum photon energy required to emit photoelectrons from the surface of $Pt$ is $E$, then that for the surface of $Ca$ would be:

• A. $2E$
• B. $1.5E$
• C. $0.5E$
• D. Cannot be determined

Answer 1

The correct option is C $0.5E$

Work function is the minimum energy required to eject an electron from the metal surface.
The work function of $Pt$ is twice that of $Ca$.
If $E$ is the minimum amount of energy required for $Pt$, then $\frac{E}{2}$ will be the minimum energy required for $Ca$.