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The de-Broglie wavelength of a bus moving 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. The de-Broglie wavelength now is:
(A) $\lambda$
(B) $2\lambda$
(C) $\dfrac{\lambda }{2}$
(D) $\dfrac{\lambda }{4}$

Last updated date: 13th Jun 2024
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Hint The de Broglie equation is defined as $\lambda = h/mv$, where $\lambda$ is the wavelength, h is the Planck’s constant, m is the mass of a particle that is moving with the velocity v. This is the concept which is used to solve the answer.

Complete step by step answer
We should know that momentum is given by:
$p = m \times v$
Now we have to put the values in the above expression to get:
$\dfrac{{\dfrac{1}{2}mv \times v}}{{\dfrac{1}{2} \times v}} = \dfrac{{2KE}}{v}$
If the Kinetic Energy as well as speed are doubled, momentum p remains unchanged.
Hence the wavelength is given as:
$\lambda = \dfrac{h}{p}$
Hence the de-Broglie wavelength will be unchanged.

The correct answer is option A.

Note We should know that matter waves are the central part of the theory of quantum mechanics, which is an example of wave particle duality. The concept of the matter behaves like a wave which was proposed by Louis de Broglie in the year 1924.
It should also be known to us that de Broglie proposed that the particles can exhibit properties of the waves.