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The group velocity of the de Broglie wav...

The group velocity of the de Broglie wave packet associated with a particle moving with velocity v is-

A

equal to v

B

less than v

C

greater than v

D

equal to speed of light

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The correct Answer is:
To find the group velocity of the de Broglie wave packet associated with a particle moving with velocity \( v \), we can follow these steps: ### Step 1: Understand the Concept of Group Velocity Group velocity is defined as the velocity at which the overall shape of the wave packet (or group of waves) propagates through space. It is given by the formula: \[ v_g = \frac{d\omega}{dk} \] where \( \omega \) is the angular frequency and \( k \) is the wave number. ### Step 2: Relate Group Velocity to Particle Velocity For a particle with a well-defined momentum \( p \), the de Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant. The momentum \( p \) can also be expressed in terms of the particle's velocity \( v \) as: \[ p = mv \] where \( m \) is the mass of the particle. ### Step 3: Calculate the Angular Frequency and Wave Number Using the de Broglie relations, we can express the wave number \( k \) as: \[ k = \frac{2\pi}{\lambda} = \frac{2\pi p}{h} = \frac{2\pi mv}{h} \] The angular frequency \( \omega \) can be related to energy \( E \) as: \[ E = \hbar \omega \] where \( \hbar = \frac{h}{2\pi} \). The energy of a particle can also be expressed as: \[ E = \frac{p^2}{2m} = \frac{(mv)^2}{2m} = \frac{mv^2}{2} \] Thus, \[ \omega = \frac{E}{\hbar} = \frac{mv^2}{2\hbar} \] ### Step 4: Differentiate to Find Group Velocity Now, we differentiate \( \omega \) with respect to \( k \): \[ \frac{d\omega}{dk} = \frac{d\left(\frac{mv^2}{2\hbar}\right)}{d\left(\frac{2\pi mv}{h}\right)} \] Since \( v \) is constant for a particle moving with velocity \( v \), we find that: \[ v_g = v \] ### Step 5: Conclusion The group velocity of the de Broglie wave packet associated with a particle moving with velocity \( v \) is: \[ v_g = v \] ### Final Answer The group velocity of the de Broglie wave packet associated with a particle moving with velocity \( v \) is \( v \). ---
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