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Chapter 6: Problem 134

The work function of a photoelectric surface is 2.00 eV. What is the maximum speed of the photoelectrons emitted from this surface when a 450 -nm light falls on it?

Short Answer

Expert verified
The maximum speed of the photoelectrons emitted from the photoelectric surface when a 450-nm light falls on it is \(1.818 × 10^6 m/s\).

Step by step solution

01

Write down the photoelectric effect equation and given data

The photoelectric effect equation is: \(E_{incident} = W + K_{max}\) Where: \(E_{incident}\) is the energy of the incident light \(W\) is the work function of the material \(K_{max}\) is the maximum kinetic energy of the emitted electrons Given data: \(W = 2.00 \, eV\) Wavelength of incident light (\(λ\)): \(450\, nm\)
02

Convert wavelength to energy

To find the energy of the incident light, we'll use the equation: \(E = \frac{hc}{λ}\) Where: \(E\) is the energy of light \(h\) is the Planck's constant (\(6.63 × 10^{−34} Js\)) \(c\) is the speed of light (\(3 × 10^8 m/s\)) \(λ\) is the wavelength of light First, convert the wavelength to meters: \(λ = 450\, nm = 450 × 10^{-9} m\) Now, calculate the energy: \(E_{incident} = \frac{6.63 × 10^{-34} Js × 3 × 10^8 m/s}{450 × 10^{-9} m}\)
03

Calculate the energy of incident light

\(E_{incident} = \frac{6.63 × 10^{-34} Js × 3 × 10^8 m/s}{450 × 10^{-9} m} = 4.42 × 10^{-19} J\) Now, convert the energy to eV: \(E_{incident} = \frac{4.42 × 10^{-19} J}{1.6 × 10^{-19} J/eV} = 2.76\, eV\)
04

Calculate the maximum kinetic energy of emitted electrons

Now that we have the energy of incident light and work function, we can find the maximum kinetic energy of emitted electrons: \(K_{max} = E_{incident} - W\) \(K_{max} = 2.76\, eV - 2.00\, eV\) \(K_{max} = 0.76\, eV\)
05

Find the maximum speed of the photoelectrons

The maximum kinetic energy \(K_{max}\) of the photoelectrons is related to their maximum speed \(v_{max}\) by the equation: \(K_{max} = \frac{1}{2} mv_{max}^2\) Where \(m\) is the mass of an electron (\(9.11 × 10^{-31} kg\)). Convert \(K_{max}\) from eV to Joules: \(K_{max} = 0.76 \, eV × 1.6 × 10^{-19} \frac{J}{eV} = 1.216 × 10^{-19} J\) Now, we can solve for \(v_{max}\): \(v_{max} = \sqrt{\frac{2K_{max}}{m}}\) \(v_{max} = \sqrt{\frac{2 × 1.216 × 10^{-19} J}{9.11 × 10^{-31} kg}}\)
06

Calculate the maximum speed

\(v_{max} = \sqrt{\frac{2 × 1.216 × 10^{-19} J}{9.11 × 10^{-31} kg}} = 1.818 × 10^6 m/s\) The maximum speed of the photoelectrons emitted from the surface when a \(450\, nm\) light falls on it is \(1.818 × 10^6 m/s\).

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