Question

Potassium and gold cathodes are used in a photoelectric effect experiment. For each cathode, find:

a) The threshold frequency.

b) The threshold wavelength.

c) The maximum photoelectron ejection speed if the light has a wavelength of 220 nm.

d) The stopping potential if the wavelength is 220 nm

Short answer

a) Threshold frequency of :

Potassium = $5.56\times {10}^{14}Hz$

Gold = $1.23\times {10}^{15}Hz$

b) Threshold wavelength of :

Potassium = $5.4\times {10}^{-7}m$

Gold = $2.44\times {10}^{-7}m$

c) The maximum photoelectron ejection speed of :

Potassium = 3.44 eV

Gold = 0.55 eV

d) Stopping potential of :

Potassium = 3.44 V

Gold = 0.55 V

Step-by-step solution

Step 1. A Potassium and a Gold Cathode is given to us.

We need to find out :

a) The threshold frequency.

b) The threshold wavelength.

c) The maximum photoelectron ejection speed if the light has a wavelength of 220 nm.

d) The stopping potential if the wavelength is 220 nm

Step 2. Part a) For Threshold Frequency

$$\begin{gathered} E_0=h{f}_0 \\ {f}_0=\frac{E_0}{h} \\ ForPotassium, \\ {f}_0=\frac{2.3eV}{4.1357\times {10}^{-15}eV.s} \\ {f}_0=5.56\times {10}^{14}Hz \\ ForGold, \\ {f}_0=\frac{5.1\cancel{eV}}{4.1357\times {10}^{-15}\cancel{eV}.s} \\ {f}_0=1.23\times {10}^{15}Hz \end{gathered}$$

Step 3. Part b) For threshold Wavelength

$$\begin{gathered} f_0=\frac{c}{{\lambda }_0} \\ {\lambda }_0=\frac{c}{f_0} \\ ForPotassium, \\ {\lambda }_0=\frac{3\times {10}^{8}m/\cancel{s}}{5.56\times {10}^{14}\cancel{s^{-1}}} \\ {\lambda }_0\simeq 5.4\times {10}^{-7}m \\ ForGold, \\ {\lambda }_0=\frac{3\times {10}^{8}m/\cancel{s}}{1.23\times {10}^{15}\cancel{s^{-1}}} \\ {\lambda }_0\simeq 2.44\times {10}^{-7}m \end{gathered}$$

Step 4. Part c) For maximum Kinetic energy of ejected electrons

$$\begin{gathered} K_{max}=E_{light}-E_0 \\ {K}_{max}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0} \\ {K}_{max}=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right) \\ ForPotassium, \\ {K}_{max}=(4.1357\times {10}^{-15}eV.s)(3\times {10}^{8}m/s)\left(\frac{1}{220\times {10}^{-9}m}-\frac{1}{5.4\times {10}^{-7}m}\right) \\ {K}_{max}=3.34eV \\ ForGold, \\ {K}_{max}=(4.1357\times {10}^{-15}eV.s)(3\times {10}^{8}m/s)\left(\frac{1}{220\times {10}^{-9}m}-\frac{1}{2.44\times {10}^{-7}m}\right) \\ {K}_{max}=0.55eV \end{gathered}$$

Step 5. Part d) Applying the Law of conservation of Energy.

$$\begin{gathered} U_B+K_B=U_A+K_A \\ AisforCathode \\ BisforAnode. \\ {U}_B=q{V}_B \\ {K}_B=0 \\ {U}_A=q{V}_A \\ {K}_A=K_{max} \\ q{V}_B+0=q{V}_A+K_{max} \\ -e{V}_B=-e{V}_A+K_{max} \\ e{V}_A-e{V}_B=K_{max} \\ e∆V=K_{max} \\ ∆V=\frac{K_{max}}{e} \\ ForPotassium, \\ ∆V=\frac{3.34eV}{e}=3.34V \\ ForGold, \\ ∆V=\frac{0.55eV}{e}=0.55V \end{gathered}$$