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Step-by-Step Solution
Step 1: Write down the de Broglie relation
The de Broglie wavelength $ \lambda $ of a particle is related to its momentum $ p $ by the equation:
$ p = \frac{h}{\lambda} $
where $ h $ is Planck’s constant.
Step 2: Determine the momentum and velocity of each electron in the lab frame
Let one electron move along the x-axis with de Broglie wavelength $ \lambda_1 $, and the other move along the y-axis with de Broglie wavelength $ \lambda_2 $. Then:
$ p_1 = \frac{h}{\lambda_1} \quad \text{(along the x-axis)} \quad,\quad
p_2 = \frac{h}{\lambda_2} \quad \text{(along the y-axis)} $
If $ m $ is the electron’s mass, the corresponding velocities in the lab frame are:
$ \vec{v}_1 = \frac{h}{m\,\lambda_1} \,\hat{i}, \quad
\vec{v}_2 = \frac{h}{m\,\lambda_2} \,\hat{j}. $
Step 3: Find the velocity of the center of mass (CM)
Since there are two electrons (each of mass $ m $), the center of mass velocity $ \vec{v}_{CM} $ is given by the average of the individual velocities:
$ \vec{v}_{CM} = \frac{\vec{v}_1 + \vec{v}_2}{2}
= \frac{1}{2}\left(\frac{h}{m\,\lambda_1}\,\hat{i} + \frac{h}{m\,\lambda_2}\,\hat{j}\right). $
Step 4: Find the velocities of each electron in the CM frame
The velocity of the first electron relative to the CM is:
$ \vec{v}_{1CM} = \vec{v}_1 - \vec{v}_{CM}
= \frac{h}{m\,\lambda_1}\,\hat{i} - \frac{1}{2}\left(\frac{h}{m\,\lambda_1}\,\hat{i} + \frac{h}{m\,\lambda_2}\,\hat{j}\right)
= \frac{1}{2}\left(\frac{h}{m\,\lambda_1}\,\hat{i} - \frac{h}{m\,\lambda_2}\,\hat{j}\right).
$
Similarly, for the second electron:
$ \vec{v}_{2CM} = \vec{v}_2 - \vec{v}_{CM}
= \frac{h}{m\,\lambda_2}\,\hat{j} - \frac{1}{2}\left(\frac{h}{m\,\lambda_1}\,\hat{i} + \frac{h}{m\,\lambda_2}\,\hat{j}\right)
= \frac{1}{2}\left(\frac{h}{m\,\lambda_2}\,\hat{j} - \frac{h}{m\,\lambda_1}\,\hat{i}\right).
$
Each of these has the same magnitude in the CM frame.
Step 5: Compute the momentum magnitude in the CM frame
The magnitude of $ \vec{v}_{1CM} $ (or $ \vec{v}_{2CM} $) is:
$ \left|\vec{v}_{1CM}\right|
= \frac{1}{2} \sqrt{ \left(\frac{h}{m\,\lambda_1}\right)^2 + \left(\frac{h}{m\,\lambda_2}\right)^2}
= \frac{h}{2m} \sqrt{ \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} }.
$
Thus, the momentum of each electron in the CM frame is:
$ p_{CM} = m \left|\vec{v}_{1CM}\right|
= \frac{h}{2} \sqrt{ \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} }.
$
Step 6: Express the de Broglie wavelength in the CM frame
Since $ p_{CM} = \frac{h}{\lambda_{CM}} $, we have:
$ \frac{h}{\lambda_{CM}}
= \frac{h}{2} \sqrt{ \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} }.
$
Canceling $ h $ on both sides gives:
$ \frac{1}{\lambda_{CM}}
= \frac{1}{2} \sqrt{ \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} }
= \frac{\sqrt{\lambda_1^2 + \lambda_2^2}} {2\,\lambda_1\,\lambda_2}.
$
Hence, the de Broglie wavelength in the CM frame is:
$ \lambda_{CM}
= \frac{2\,\lambda_1\,\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}.
$
Final Answer
$ \displaystyle \lambda_{CM}
= \frac{2\,\lambda_1\,\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}.
$
