Photoemission calculations - worked examples
Question
A photon has a frequency of \(6.4 \times 10^{14}Hz\). Calculate the energy of the photon.
\(h=6.63 \times 10^{-34}\)
\(f = 6.4 \times 10^{14}\)
\(E = hf\)
\(E = 6.4 \times 10^{14} \times 6.63 \times 10^{-34}\)
\(E = 4.24 \times 10^{-19} J\)
The energy of the photon is \(4.24 \times 10^{-19} J\)
Question
The work function of sodium metal is \(3.6\times 10^{-19}J\).
Calculate the minimum frequency of a photon which will cause photo emission.
\(W = 3.45 \times 10^{-19}\)
\(h = 6.63 \times 10^{-34}\)
\(f_{o} = \frac{W}{h}\)
\(f_{o}=\frac{3.6\times 10^{-19}}{6.63\times 10^{-34}}\)
\(f_{o} = 5.43 \times 10^{14}Hz\)
The minimum (threshold) frequency is \(5.43 \times 10^{14}Hz\)
Question
Now calculate the wavelength of this photon.
Photons travel at the speed of light.
\(v = 3 \times 10^{8}ms^{-1}\)
\(f = 5.43 \times 10^{14}Hz\)
\(\lambda = \frac{v}{f}\)
\(\lambda =\frac{3\times 10^{8}}{5.43\times 10^{14}}\)
\(\lambda= 5.52\times 10^{-7} m\)
\(\lambda= 552 nm\)
The wavelength of the photon is \(552 nm\).
Question
Light of frequency \(6.4\times 10^{14}Hz\) is used to illuminate negatively charged sodium (which has a work function of \(3.6\times 10^{-19}J\)).
Calculate the speed of emitted electrons if the mass of an electron is \(9.11\times 10^{-31}kg\).
\(E_{k}= E-W\)
\(= (4.24\times 10^{-19})-(3.45\times 10^{-19})\)
\(=7.9\times10^{-20}J\)
This equals the kinetic energy of the electron.
Remember that \(E_{k}=\frac{1}{2} mv^{2}\)
\(\frac{1}{2}mv^{2}=7.9\times10^{-20}\)
\(0.5\times9.11\times10^{-31}v^{2}=7.9\times10^{-20}\)
\(v^{2}=(7.9\times10^{-20})\div(0.5\times9.11\times10^{-31})\)
\(v^{2}=1.73\times10^{11}\)
\(v=\sqrt{1.73\times10^{11}}\)
\(v=4.16\times10^{5}ms^{-1}\)