Question

Metal 1 has a larger work function than metal 2. Both are illuminated with the same short-wavelength ultraviolet light. Do photoelectrons from metal 1 have a higher speed, a lower speed, or the same speed as photoelectrons from metal 2? Explain.

Short answer

Photoelectrons from metal 1 will have a lower speed than metal 2.

Step-by-step solution

Step 1. Two metals 1 and 2 are illuminated with same short wavelength and :Work function of metal 1 > Work function of metal 2

We need to find out whether photoelectrons of metal 1 will have higher, lower or equal speed to that to photoelectrons of metal 2.

Step 2. Comparing the speed of metal 1 and metal 2.

$$\begin{gathered} Since{\lambda }_1={\lambda }_2 \\ {f}_1=f_2 \\ And{\left(E_{elec}\right)}_1=h{f}_{1}isequalto{\left(E_{elec}\right)}_2=h{f}_2 \\ But{\left(E_0\right)}_1={\left(E_0\right)}_2 \\ Therefore, \\ {\left(K_{max}\right)}_1=E_{elec}-{\left(E_0\right)}_1 \\ {\left(K_{max}\right)}_2=E_{elec}-{\left(E_0\right)}_2 \\ So, \\ {\left(K_{max}\right)}_1{<\left(K_{max}\right)}_2 \\ Hence, \\ {v}_1<v_2 \end{gathered}$$

Photoelectrons from metal 1 will have a lower speed than metal 2.