Question

Two photons having
(A) equal wavelength have equal linear momenta
(B) equal energies have equal linear momenta
(C) equal frequencies have equal linear momenta
(D) equal linear momenta have equal wavelengths

Answer 1

Hint : Use De Brogile’s equation and equation for energy of a photon, and the definition of linear momentum to check each of the options. De Broglie’s wavelength is given by, $\lambda = \dfrac{h}{p}$ where $p$ is the linear momentum of the wave /particle, $h$ is the Planck’s constant. Energy of a wave with frequency $\nu$ is, $E = h\nu$

Complete Step By Step Answer:
We know the de Broglie’s wavelength of a wave ( particle) dual is given by, $\lambda = \dfrac{h}{p}$ where $h$ is the Planck’s constant, $p$ is the linear momentum of the wave /particle.
Now, to solve this problem, we can check each option one by one and eliminate one by one.
So, let’s check option A first, if both the photons have equal wavelength then we can have, ${\lambda _1} = {\lambda _2}$
Or, $\dfrac{h}{{{p_1}}} = \dfrac{h}{{{p_2}}}$
Hence, ${p_1} = {p_2}$
But we know, momentum is a vector quantity given by, $\vec p = m\vec v$where $m$is the mass of the body.(for a classical particle), for a wave it is $p = \dfrac{h}{\lambda }$direction of it is along the direction of propagation $\vec k$.
Here, we have the magnitude of momentum is same, but the direction of the momentum can be different .Hence option (a) is incorrect
Let’s Check option (b),
We have here photon with equal energies, that is, ${E_1} = {E_2}$
or, $h{\nu _1} = h{\nu _2}$
Now, $p = \dfrac{{h\nu }}{c}$
Therefore we have, $h\dfrac{{c{p_1}}}{h} = h\dfrac{{c{p_2}}}{h}$
Or, ${p_1} = {p_2}$
Again we can say that, if also the magnitude of linear momentum is equal the direction of it can be different.
Hence, option (b) is also incorrect.
Let’s Check option (C) then,
We have here, photon of equal frequency, ${\nu _1} = {\nu _2}$ again putting the value of frequency in terms of linear momentum we get,
$\dfrac{{c{p_1}}}{h} = \dfrac{{c{p_2}}}{h}$
Or, ${p_1} = {p_2}$
Again we can say that, though the magnitude of linear momentum is equal the direction of it can be different.
Hence, option (C ) is also incorrect.
Let’s check option (D),
Here we have a photon of equal linear momentum, ${\vec p_1} = {\vec p_2}$. That means the direction of them must also be.
Now, taking only magnitude and putting the value of momentum in terms of wavelength we get,
$\dfrac{h}{{{\lambda _1}}} = \dfrac{h}{{{\lambda _2}}}$
Or, ${\lambda _1} = {\lambda _2}$
Therefore, the wavelength of the photons is equal.
Hence, option (D) is correct.

Note :
The wavelength of a wave with particle nature is given by de Broglie’s wavelength. The equation is a scalar one, not a vector quantity. So, provided the equation we can find the wave properties such as wavelength, frequency energy etc. but not the direction of propagation. To find the direction of the propagation we need a wave function with boundary conditions. The general form of a normalized wave function is given by, $\Psi = A{e^{ - i(\vec k.\vec r - \omega t)}}$ . $A$is the normalization constant, $\vec k$propagation constant, $\vec r$ is the position vector, $\omega$ is the frequency of the wave.