Question

If we assume kinetic energy of a proton is equal to energy of the photon, the ratio of de Broglie wavelength of proton to photon is proportional to
A. E
B. $E^{-1/2}$
C. $E^{1/2}$
D. $E^{3/2}$

Answer 1

Hint: As a photon has zero rest mass, so its energy is just a product of momentum and c. A proton's kinetic energy is the difference of its total energy and rest mass energy. De Broglie wavelength is related to the momentum of the particle.

Formula used:
The de Broglie wavelength associated with a particle with momentum p is:
$\lambda = \dfrac{h}{p}$.

Complete step-by-step answer:
When a particle moves at relativistic velocities, it's total energy is written as:
$E = \sqrt{ p^2 c^2 + m_0^2 c^4}$
where $m_0$ happens to be the rest mass for the particle and p is the momentum of the particle.
The kinetic energy for such particle is written as
$K = E - m_0 c^2$
i.e., if we subtract rest mass energy from total energy we get kinetic energy.

To simplify the mathematics involved we assume that the photon in question is a low energy photon and the proton also has low energy. Therefore for low energies we can write:
$\dfrac{p^2}{2 m_0} = K \implies p = \sqrt{2 m_0 K}$.
The de Broglie wavelength associated with proton can be written as:
$\lambda' = \dfrac{h}{\sqrt{2 m_0 K}}$.

Now, for a photon, rest mass is zero, so it's total energy is it's kinetic energy which is:
$E = pc$
which gives us the de Broglie wavelength:
$\lambda = \dfrac{h c}{E}$ .

We can write the ratio as:
$\dfrac{\lambda'}{\lambda } = \dfrac{h / \sqrt{2 m_0 K}}{ hc / E} = \dfrac{ E }{\sqrt{2 m_0 c^2 K} } $.
Now, we are given K = E as kinetic energy of both proton and photon is same,
$\dfrac{\lambda'}{\lambda } = \sqrt{ \dfrac{ E }{2 m_0 c^2 } } $.
This suggests that the ratio is directly proportional to $E^{1/2}$.

So, the correct answer is “Option C”.

Note: If the particle in question has high energy, then we cannot use the same expression for kinetic energy as we did here. It is recommended that one should carry out all the calculations considering the kinetic energy for a relativistic particle at high energies too. Remember that our usual kinetic energy relation is not good at high energies or relativistic velocities.