Question

Given that the work function of tungsten is \(4.55 \mathrm{eV}\), what is the stopping potential in an experiment using tungsten cathodes at \(360 \mathrm{nm} ?\)

Short answer

Answer: The stopping potential in the experiment using tungsten cathodes at 360 nm is +1.105 V.

Step-by-step solution

List down the given information

We are given the following information: Work function of tungsten: \(4.55 \mathrm{eV}\) Wavelength of light incident on the cathode: \(360 \mathrm{nm}\)

Convert wavelength to energy

The energy of the incident light can be found using Planck's Law, which relates the energy of a photon to its wavelength: \(E = \dfrac{h \cdot c}{\lambda}\) where \(E\) is the energy of the incident light, \(h\) is Planck's constant (\(6.62607 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\)), \(c\) is the speed of light (\(3 \times 10^8 \mathrm{m/s}\)), and \(\lambda\) is the wavelength. First, convert the wavelength from nanometers to meters: \(360 \mathrm{nm} = 360 \times 10^{-9} \mathrm{m}\) Now, calculate the energy: \(E = \dfrac{(6.62607 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}) \cdot (3 \times 10^8 \mathrm{m/s})}{360 \times 10^{-9} \mathrm{m}}\) \(E = 5.519751 \times 10^{-19} \mathrm{J}\) Since we need the energy in electron-volts (eV), we need to convert it using the relation: \(1 \mathrm{eV} = 1.60218 \times 10^{-19} \mathrm{J}\): \(E = \dfrac{5.519751 \times 10^{-19} \mathrm{J}}{1.60218 \times 10^{-19} \mathrm{J/eV}}\) \(E = 3.445 \mathrm{eV}\)

Calculate the stopping potential using the photoelectric effect equation

The photoelectric effect equation can be written as: \(E = (W + V_s \cdot e)\) where \(E\) is the energy of the incident light, \(W\) is the work function of the material, \(V_s\) is the stopping potential, and \(e\) is the charge of the electron (also written as \(1.60218 \times 10^{-19} \mathrm{C}\)). Substitute the known values and solve for \(V_s\): \(3.445 \mathrm{eV} = (4.55 \mathrm{eV} + V_s \cdot e)\) To find the stopping potential, first, subtract the work function from the energy of the incident light: \(V_s \cdot e = 3.445 \mathrm{eV} - 4.55 \mathrm{eV}\) \(V_s \cdot e = -1.105 \mathrm{eV}\) Now, divide by the charge of the electron to find the stopping potential: \(V_s = \dfrac{-1.105 \mathrm{eV}}{e}\) Since \(V_s\) is negative, the stopping potential will be positive (since the potential is applied in the opposite direction to stop the electrons): \(V_s = +1.105 \mathrm{V}\)

State the final answer

The stopping potential in the experiment using tungsten cathodes at \(360 \mathrm{nm}\) is \(+1.105 \mathrm{V}\).