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:

For any particle moving non-relativistically, the de Broglie relation states

$$\lambda=\dfrac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle.

Let the magnitudes of the momenta of the two electrons in the laboratory frame be $$p_1$$ and $$p_2$$. Using the above relation we can write

$$p_1=\dfrac{h}{\lambda_1}, \qquad p_2=\dfrac{h}{\lambda_2}.$$

We are told that the velocity vectors of the two electrons are perpendicular to each other. We can choose a convenient set of axes so that

$$\vec p_1=p_1\,\hat i, \qquad \vec p_2=p_2\,\hat j.$$

The total momentum of the two-electron system is therefore the vector sum

$$\vec P_{\text{tot}}=\vec p_1+\vec p_2=p_1\,\hat i+p_2\,\hat j.$$

Its magnitude is obtained from the Pythagorean theorem because the two components are perpendicular:

$$|\vec P_{\text{tot}}|=\sqrt{p_1^{\,2}+p_2^{\,2}}.$$

The centre-of-mass (CM) velocity of the system is, by definition,

$$\vec V_{\!CM}=\dfrac{\vec P_{\text{tot}}}{m_1+m_2}.$$

Both particles are electrons, so $$m_1=m_2=m_e$$, giving

$$\vec V_{\!CM}=\dfrac{\vec P_{\text{tot}}}{2m_e} =\dfrac{p_1}{2m_e}\,\hat i+\dfrac{p_2}{2m_e}\,\hat j.$$

Now we determine the momentum of, say, electron 1 in the CM frame. The Galilean transformation for momenta at non-relativistic speeds is

$$\vec p_1'=\vec p_1-m_e\vec V_{\!CM}.$$

Substituting the expressions for $$\vec p_1$$ and $$\vec V_{\!CM}$$, we get

$$\vec p_1'= p_1\,\hat i- m_e\!\left(\dfrac{p_1}{2m_e}\,\hat i+\dfrac{p_2}{2m_e}\,\hat j\right) =\Bigl(p_1-\dfrac{p_1}{2}\Bigr)\hat i-\dfrac{p_2}{2}\hat j =\dfrac{p_1}{2}\,\hat i-\dfrac{p_2}{2}\,\hat j.$$

The magnitude of this vector is therefore

$$p_{\!CM}=|\vec p_1'|=\sqrt{\left(\dfrac{p_1}{2}\right)^{\!2}+ \left(\dfrac{p_2}{2}\right)^{\!2}} =\dfrac{1}{2}\sqrt{p_1^{\,2}+p_2^{\,2}}.$$

Because the two electrons have equal mass, electron 2 possesses the same magnitude of momentum in the CM frame; hence the common CM-frame de Broglie wavelength is

$$\lambda_{CM}=\dfrac{h}{p_{\!CM}} =\dfrac{h}{\dfrac{1}{2}\sqrt{p_1^{\,2}+p_2^{\,2}}} =\dfrac{2h}{\sqrt{p_1^{\,2}+p_2^{\,2}}}.$$

Next we eliminate the momenta in favour of the given wavelengths. Recalling $$p_1=h/\lambda_1$$ and $$p_2=h/\lambda_2$$, we have

$$p_1^{\,2}+p_2^{\,2} =\left(\dfrac{h}{\lambda_1}\right)^{\!2}+\left(\dfrac{h}{\lambda_2}\right)^{\!2} =h^{2}\!\left(\dfrac{1}{\lambda_1^{\,2}}+\dfrac{1}{\lambda_2^{\,2}}\right).$$

Inserting this into the expression for $$\lambda_{CM}$$ gives

$$\lambda_{CM} =\dfrac{2h} {h\sqrt{\dfrac{1}{\lambda_1^{\,2}}+\dfrac{1}{\lambda_2^{\,2}}}} =\dfrac{2} {\sqrt{\dfrac{1}{\lambda_1^{\,2}}+\dfrac{1}{\lambda_2^{\,2}}}}.$$

To put everything over a common denominator, write

$$\dfrac{1}{\lambda_1^{\,2}}+\dfrac{1}{\lambda_2^{\,2}} =\dfrac{\lambda_2^{\,2}+\lambda_1^{\,2}}{\lambda_1^{\,2}\lambda_2^{\,2}}.$$

Therefore

$$\sqrt{\dfrac{1}{\lambda_1^{\,2}}+\dfrac{1}{\lambda_2^{\,2}}} =\dfrac{\sqrt{\lambda_1^{\,2}+\lambda_2^{\,2}}}{\lambda_1\lambda_2}.$$

Substituting this result into the previous expression for $$\lambda_{CM}$$, we finally reach

$$\lambda_{CM} =\dfrac{2}{\dfrac{\sqrt{\lambda_1^{\,2}+\lambda_2^{\,2}}}{\lambda_1\lambda_2}} =\dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^{\,2}+\lambda_2^{\,2}}}.$$

This matches option C.

Hence, the correct answer is Option C.