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PhysicsModern PhysicsDe Broglie & Matter WavesMedium2 minPYQ_2018

PhysicsMediumstatement

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de brogile wavelengths are $λ_{1}$ and $λ_{2}$ , their de brogile wavelength in the frame of reference attached to their centre of mass is:

Options:

Answer:

Solution:

Momentum of each electron $\frac{h}{λ_{1}} \hat{i} a n d \frac{h}{λ_{2}} \hat{j}$

Velocity of centre of mass

$V_{c m} = \frac{h}{2 m λ_{1}} \hat{i} + \frac{h}{2 m λ_{2}} \hat{j}$

Velocity of $1^{s t}$ particle about centre of mass

$V_{c m} = \frac{h}{2 m λ_{1}} \hat{i} - \frac{h}{2 m λ_{2}} \hat{j}$

$λ_{c m} = \frac{h}{\sqrt{\frac{h^{2}}{4 λ_{1}^{2}} + \frac{h^{2}}{4 λ_{2}^{2}}}} = \frac{2 λ_{1} λ_{2}}{\sqrt{λ_{1}^{2} + λ_{2}^{2}}}$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:De Broglie & Matter Waves

⏱ 2mℹ️ Source: PYQ_2018

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