 # JC-23) de Broglie Waves A photon and a particle have the same wavelength. a) How do their linear momenta compare? b) How does the photon's energy compare to the particle's total energy? c) How does the photon's energy compare to the particle's kinetic energy? przesypkai4 2022-07-16 Answered
JC-23) de Broglie Waves A photon and a particle have the same wavelength.
a) How do their linear momenta compare?
b) How does the photon's energy compare to the particle's total energy?
c) How does the photon's energy compare to the particle's kinetic energy?
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Given : A photon have wavelength
Particle wave length
To find : a) How do their linear momenta compare?
b) How does the photon's energy compare to the particle's total energy?
c) How does the photon's energy compare to the particle's kinetic energy?
Solution : Any particle’s wavelength is determined by its momentum, and hence particles with the same wavelength have the same momenta. Acc. to de Broglie's equation,
$p=\frac{h}{\lambda }$
p is momentum
h is planck's constant
$\lambda$ is wavelength
wavelength is same for particle and photon, So linear momenta is also same.
b) The energy of photon is given by,
$E=pc..................\left(1\right)$
E is energy
c is speed of light
The energy of particle is given by,
$E=\sqrt{\left(pc{\right)}^{2}+\left(mc{\right)}^{2}}................\left(2\right)$
from equation (1) and (2) its clear that particle energy is greater than photon.
c) Kinetic energy of photon is given by.
$K\cdot E=pc$ ...............(3)
In case of particle,
For non relativistic particle
$K\cdot E=\frac{{p}^{2}}{2m}$ .....................(4)
For relativistic particle.
$K.E=\sqrt{\left(pc{\right)}^{2}+\left(m{c}^{2}{\right)}^{2}}-m{c}^{2}$
In both the cases of particle (relativistic and non-relativistic ) kinetic energy is less than kinetic energy of photon.
You can compare mathematically equation 3, 4 and 5.