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(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Initially, let the mass be $m$, speed be $v$, and kinetic ene

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$\lambda$

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(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Initially, let the mass be $m$, speed be $v$, and kinetic energy be $E = \frac{1}{2}mv^2$. Thus, $\lambda = \frac{h}{\sqrt{2mE}}$.After some passengers leave, let the new mass be $m'$, new speed be $v' = 2v$, and new kinetic energy be $E' = 2E$.Using the kinetic energy formula: $E' = \frac{1}{2}m'(v')^2$.Substituting the values: $2E = \frac{1}{2}m'(2v)^2 = \frac{1}{2}m'(4v^2) = 2m'v^2$.Since $E = \frac{1}{2}mv^2$, we have $2(\frac{1}{2}mv^2) = 2m'v^2$, which implies $m' = \frac{m}{2}$.Now, the new de-Broglie wavelength $\lambda'$ is:$\lambda' = \frac{h}{\sqrt{2m'E'}} = \frac{h}{\sqrt{2(\frac{m}{2})(2E)}} = \frac{h}{\sqrt{2mE}} = \lambda$.Therefore, the new de-Broglie wavelength remains $\lambda$.

The de-Broglie wavelength of a moving bus with speed $v$ is $\lambda$. Some passengers left the bus at a stoppage. Now, when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. What is the de-Broglie wavelength of the bus now?

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