Question

The de-Broglie wavelength of a bus moving with speed 'v' is ${}^{\prime }{\lambda }^{\prime }$ .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. $\lambda /2$
• D. $\lambda /4$

Answer 1

The correct option is A $\lambda$

Initially

$\lambda =\frac{h}{mv}$

Now

$\frac{1}{2}M(2v{)}^2=2\times \frac{1}{2}m{v}^2$

$2M{v}^2=m{v}^2$

$M=\frac{m}{2}$

$so{\lambda }_{Now}=\frac{h}{\frac{m}{2}\times 2v}$
$=\lambda$

So, the answer is option (A).