Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are \lambda _1 and \lambda _2, their de Broglie wavelength in the frame of reference attached to their center of masses :

\lambda _CM = 2\lambda _1\lambda _2 \over \sqrt \lambda _1^2 + \lambda _2^2

\lambda _CM = \lambda _1 = \lambda _2

1 \over \lambda _CM = 1 \over \lambda _1 + 1 \over \lambda _2

\lambda _CM = ( \lambda _1 + \lambda _2 \over 2 )

Dual Nature of Radiation PYQ — JEE Main 2018 · 15 Apr · Shift 1 · Q55 with Solution

View written solutionFree

Correct answer: B

Relate de Broglie wavelength to momentum

For a non-relativistic particle,

\lambda = \frac{h}{p}

So for the two electrons,

p_1 = \frac{h}{\lambda_1}, \qquad p_2 = \frac{h}{\lambda_2}

Use the fact that the electrons move perpendicular to each other

Since the momenta are perpendicular,

|\vec p_1 + \vec p_2| = \sqrt{p_1^2 + p_2^2}

Velocity of the center of mass frame

Both particles are electrons, so their masses are equal ( $m$ ). In the center of mass frame, each particle has momentum equal in magnitude to half the total momentum of the system subtracted appropriately.

For two equal masses, the momentum of either electron in the CM frame is

p_{CM} = \left|\vec p_1 - \frac{\vec p_1+\vec p_2}{2}\right| = \frac{1}{2}|\vec p_1 - \vec p_2|

Since $\vec p_1 \perp \vec p_2$ ,

|\vec p_1 - \vec p_2| = \sqrt{p_1^2 + p_2^2}

Hence,

p_{CM} = \frac{1}{2}\sqrt{p_1^2 + p_2^2}

Find the de Broglie wavelength in the CM frame

Therefore,

\lambda_{CM} = \frac{h}{p_{CM}} = \frac{h}{\frac{1}{2}\sqrt{p_1^2+p_2^2}} = \frac{2h}{\sqrt{p_1^2+p_2^2}}

Substitute $p_1=\frac{h}{\lambda_1}$ and $p_2=\frac{h}{\lambda_2}$ :

\lambda_{CM} = \frac{2h}{\sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2}}

Factor out $h$ :

\lambda_{CM} = \frac{2}{\sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}}}

Now simplify:

\lambda_{CM} = \frac{2}{\sqrt{\frac{\lambda_1^2+\lambda_2^2}{\lambda_1^2\lambda_2^2}}} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}

Match with the options

This is exactly:

\boxed{\lambda_{CM} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}}

So the correct option is B.

More from Dual Nature of Radiation

View written solutionFree

Correct answer: B

Relate de Broglie wavelength to momentum

For a non-relativistic particle,

\lambda = \frac{h}{p}

So for the two electrons,

p_1 = \frac{h}{\lambda_1}, \qquad p_2 = \frac{h}{\lambda_2}

Use the fact that the electrons move perpendicular to each other

Since the momenta are perpendicular,

|\vec p_1 + \vec p_2| = \sqrt{p_1^2 + p_2^2}

Velocity of the center of mass frame

For two equal masses, the momentum of either electron in the CM frame is

p_{CM} = \left|\vec p_1 - \frac{\vec p_1+\vec p_2}{2}\right| = \frac{1}{2}|\vec p_1 - \vec p_2|

Since $\vec p_1 \perp \vec p_2$ ,

|\vec p_1 - \vec p_2| = \sqrt{p_1^2 + p_2^2}

Hence,

p_{CM} = \frac{1}{2}\sqrt{p_1^2 + p_2^2}

Find the de Broglie wavelength in the CM frame

Therefore,

\lambda_{CM} = \frac{h}{p_{CM}} = \frac{h}{\frac{1}{2}\sqrt{p_1^2+p_2^2}} = \frac{2h}{\sqrt{p_1^2+p_2^2}}

Substitute $p_1=\frac{h}{\lambda_1}$ and $p_2=\frac{h}{\lambda_2}$ :

\lambda_{CM} = \frac{2h}{\sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2}}

Factor out $h$ :

\lambda_{CM} = \frac{2}{\sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}}}

Now simplify:

\lambda_{CM} = \frac{2}{\sqrt{\frac{\lambda_1^2+\lambda_2^2}{\lambda_1^2\lambda_2^2}}} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}

Match with the options

This is exactly:

\boxed{\lambda_{CM} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}}

So the correct option is B.

Dual Nature of Radiation question

Dual Nature of Radiation question

More from Dual Nature of Radiation

Dual Nature of Radiation question

Dual Nature of Radiation question

More from Dual Nature of Radiation