91影视

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

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

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

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

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