The wavelength of a very fast-moving electron $(v \approx c)$ is:

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is:

For an electron microscope,which of the following is false?

$A$ proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is $2000$ times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(\lambda_{p})$ and that of the electron $(\lambda_{e})$?

$A$ proton and an $\alpha$-particle are accelerated through a potential difference of $100\, V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha$-particle is

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is