Question

An electron has a kinetic energy of 3.00 ev. find its wavelength. (b) what if? a photon has energy 3.00 ev. find its wavelength.

Answer 1

(a) The electron kinetic energy is
$K=3.00 eV$
which can be converted into Joule by keeping in mind that
$1 eV=1.6 \cdot 10^{-19}eV$
So that we find
$K=3.00 eV \cdot 1.6 \cdot 10^{-19} eV/J =4.8 \cdot 10^{-19}J$

The kinetic energy of the electron is related to its momentum p by:
$K= \frac{p^2}{2m}$
where m is the electron mass. Re-arranging the equation, we find
$p= \sqrt{ 2Km}= \sqrt{ 2 ( 4.8 \cdot 10^{-19} J)(9.1 \cdot 10^{-31} kg) } =9.35 \cdot 10^{-25} kgm/s$

And now we can use De Broglie's relationship to find its wavelength:
$\lambda= \frac{h}{p}= \frac{6.6 \cdot 10^{-34} Js}{9.35 \cdot 10^{-25} kg m/s} =7.06 \cdot 10^{-10}m$
where h is the Planck constant.


(b) By using the same procedure of part (a), we can convert the photon energy into Joules:
$E=3.00 eV \cdot 1.6 \cdot 10^{-19} eV/J =4.8 \cdot 10^{-19}J$

The energy of a photon is related to its frequency f by:
$E=hf$
where h is the Planck constant. Re-arranging the equation, we find
$f= \frac{E}{h}= \frac{4.8 \cdot 10^{-19} J}{6.6 \cdot 10^{-34}Js} =7.27 \cdot 10^{14}Hz$

And now we can use the relationship between frequency f, speed of light c and wavelength $\lambda$ of a photon, to find its wavelength:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.27 \cdot 10^{14} Hz} =4.13 \cdot 10^{-7} m$