Question

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are ${\lambda _1}$ and ${\lambda _2}$, their de Broglie wavelength in the frame of reference attached to their centre of mass is:
A. ${\lambda _{CM}} = {\lambda _1} = {\lambda _2}$
B. $\dfrac{1}{{{\lambda _{CM}}}} = \dfrac{1}{{{\lambda _1}}} + \dfrac{1}{{{\lambda _2}}}$
C. ${\lambda _{CM}} = \dfrac{{2{\lambda _1}{\lambda _2}}}{{\sqrt {{\lambda _1}^2 + {\lambda _2}^2} }}$
D. ${\lambda _{CM}} = \left( {\dfrac{{{\lambda _1} + {\lambda _2}}}{2}} \right)$

Answer 1

For solving this problem, it is obvious that we need to use the de Broglie wavelength equation. But, to find the wavelength at the centre of mass, first we have to find the velocity at the centre of mass and by using that we can find the wavelength at the centre of mass.

Formula used:
De Broglie wavelength
$\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}$
where, $\lambda$is De Broglie’s wavelength
$h$is the Planck's constant
$p$is momentum
$v$ is velocity of the particle

Complete step by step solution:
Here, we are given that both electrons are moving in perpendicular direction.Therefore, the momentum of both electrons can be taken as:

$${p_1} = \dfrac{h}{{{\lambda _1}}}\widehat i \\\ \Rightarrow{p_2} = \dfrac{h}{{{\lambda _2}}}\widehat j \\\$$

Now, we will find the velocity at the centre of mass which is given by
$\overrightarrow {{v_{CM}}} = \dfrac{{\overrightarrow {{p_1}} + \overrightarrow {{p_2}} }}{M}$
Here, $M$is the total mass of electrons. Let us consider the mass of a single electron as $m$.
Therefore, $M = 2m$
Putting all the values, we get

\overrightarrow {{v_{CM}}} = \dfrac{{m\overrightarrow {{v_1}} + m\overrightarrow {{v_2}} }}{{2m}} \\\ \Rightarrow \overrightarrow {{v_{CM}}} = \dfrac{{\overrightarrow {{v_1}} +\overrightarrow {{v_2}} }}{2}$$ Now, velocity of the first electron with respect to centre of mass. $ \overrightarrow {{v_{1CM}}} = \overrightarrow {{v_1}} - \overrightarrow {{v_{CM}}} \\\ \Rightarrow \overrightarrow {{v_{1CM}}} = \overrightarrow {{v_1}} - \dfrac{{\overrightarrow {{v_1}} + \overrightarrow {{v_2}} }}{2} = \dfrac{{\overrightarrow {{v_1}} - \overrightarrow {{v_2}} }}{2} $ Now the magnitude of this velocity is similar for both the electrons. $$\left| {\overrightarrow {{V_{1CM}}} } \right| = \sqrt {\dfrac{{{{\left| {{v_1} - {v_2}} \right|}^2}}}{2}} = \left| {\overrightarrow {{V_{2CM}}} } \right|$$ From de Broglie’s wavelength equation, we can say that their de Broglie wavelength in the frame of reference attached to their centre of mass is, $ {\lambda _{CM}} = \dfrac{h}{{m\left| {\overrightarrow {{v_{1CM}}} } \right|}} = \dfrac{h}{m} \times \dfrac{2}{{\dfrac{h}{m}\sqrt {\dfrac{1}{{{\lambda _1}^2}} + \dfrac{1}{{{\lambda _2}^2}}} }} \\\ \therefore {\lambda _{CM}} = \dfrac{{2{\lambda _1}{\lambda _2}}}{{\sqrt {{\lambda _1}^2 + {\lambda _2}^2} }} $ **Hence, option C is the right answer.** **Note:** It is important to know that In the case of electrons going in circles around the nuclei in atoms, the de Broglie waves exist as a closed-loop, such that they can exist only as standing waves, and fit evenly around the loop. This is why the electrons in atoms circle the nucleus in particular configurations, or states, which are called stationary orbits.