(a) for a photon, the relationship between frequency f and energy E is
[tex]E=hf[/tex] (1)
where h is the Planck cosntant.
First, we need to convert the energy from eV to Joule:
[tex]E=5.40 eV \cdot (1.6 \cdot 10^{-19}J/eV)=8.6 \cdot 10^{-19} J[/tex]
Then, using (1) we can find the frequency of the photon:
[tex]f= \frac{E}{h}= \frac{8.6 \cdot 10^{-19} J}{6.6 \cdot 10^{-34}Js}=1.3 \cdot 10^{15} Hz [/tex]
and finally we can find its wavelength:
[tex]\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.3 \cdot 10^{15}Hz} = 2.3 \cdot 10^{-7}m[/tex]
(b) now we have to find the De Broglie wavelength of an electron with same energy of the previous photon. Again, we must convert the energy in Joules first:
[tex]E=5.40 eV \cdot (1.6 \cdot 10^{-19}J/eV)=8.6 \cdot 10^{-19} J[/tex]
Then we can use the relationship between momentum p and energy E of a particle, to find p: (electron mass: [tex]m_e = 9.1 \cdot 10^{-31} kg[/tex] )
[tex]E= \frac{p^2}{2m} [/tex]
[tex]p= \sqrt{2Em}= \sqrt{2 (8.6 \cdot 10^{-19} J)(9.1 \cdot 10^{-31} kg)} =1.3 \cdot 10^{-24} kg m/s [/tex]
And finally we can use De Broglie relationship to find the wavelength of the electron:
[tex]\lambda = \frac{h}{p} = \frac{6.6 \cdot 10^{-34}Js}{1.4 \cdot 10^{-24} kg m/s}=5.3 \cdot 10^{-10} m [/tex]
