The de-Broglie wavelength of a moving bus wit | vedclass

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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

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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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