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

Options

A. `lambda`
B. `2lambda`
C. `lambda//2`
D. `lambda//4`

Answer

A

Explanation

Momentum,
`p=mv=((1)/(2)mvxxv)/((1)/(2)xxv)=(2KE)/(v)`
If KE as well as speed are doubled, momentum p remains unchanged.
`because " "lambda=(h)/(p)`
Hence, de-Brogile wavelength will be unchanged.