## Related questions with answers

Consider a freely moving quantum particle with mass m and speed v. Its energy is $E=K+U=\frac{1}{2} m v^{2}+0$. Determine the phase speed of the quantum wave representing the particle and show that it is different from the speed at which the particle transports mass and energy.

Solution

VerifiedAs we know, the particle has energy :

$\begin{align*} E&=K+U\\ E&=\frac{1}{2}mv^2+0\\ E&=K=\frac{1}{2}mv^2\\ E&=hf \\ \Rightarrow \lambda&=\frac{h}{mv} \tag{Wavelength.}\\ \Rightarrow v_{p}&=f \lambda \tag{Phase velocity $\rightarrow$ equation 1.}\\ \end{align*}$

Now, we have to substitute values in equation 1 and solve this equation :

$\begin{align*} v_{p}&=f \lambda \tag{Equation 1.}\\ v_{p}&=\frac{E}{h} \cdot \frac{h}{mv} \\ v_{p}&=\frac{\frac{1}{2} mv^2}{h} \cdot \frac{h}{mv} \tag{Substitute values in equation.}\\ v_{p}&=\frac{mv^2}{2h} \cdot \frac{h}{mv} \\ v_{p}&=\frac{v}{2} \ne v \\\ \end{align*}$

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