Light of wavelength 3.0 × 10−7 m shines on the metals lithium, iron, and mercury

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?

Expert Solution

\(356-23-17 R\)

(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

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

Dr. OWL answered 3 years ago

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