Question

Light of wavelength 3.0 × 10⁻⁷ m shines on the metals lithium, iron, and mercury, which have work functions of 2.3 eV, 3.9 eV, and 4.5 eV, respectively.

• a. Which of these metals will exhibit the photoelectric effect?
• b. For those metals that do exhibit the photoelectric effect, what is the maximum kinetic energy of the photoelectrons?

Answer 1

\(356-23-17 R\)

(a) Here \(\lambda=3 \times 10^{-7} \mathrm{~m}\)

Hence energy of light

\(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-7}}\)

\(\mathrm{E}=6.63 \times 10^{-19} \mathrm{Joule}\)

Now conversion to electron volts we have

\(\mathrm{E}=\frac{6.63 \times 10^{-19}}{1.6 \times 10^{-19}}\)

\(=\frac{6.63}{1.6}\)

\(=(4.14) \mathrm{eV}\)

Hence Lithium \& Iron will show photoelectric effect.

 

(b) For Lithium we have

\(\begin{aligned}(K . E)_{\max } &=4.14-2.3 \\ &=1.84 \mathrm{eV} \end{aligned}\)

For Iron

\(\begin{aligned}(K . E)_{\max } &=4.14-3.9 \\ &=0.24 \mathrm{eV} \end{aligned}\)