An electron is moving with initial velocity $\vec{v} = V_0 \hat{j}$ in a magnetic field $\vec{B} = B_0 \hat{i}$. Its de-Broglie wavelength will

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$-axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$. If both particles move along the $x$-axis after the collision,the change $\Delta \lambda$ in the de-Broglie wavelength of particle $A$,in terms of its de-Broglie wavelength $(\lambda_{0})$ before the collision is:

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

What is the de Broglie wavelength of
$(a)$ a bullet of mass $0.040 \; kg$ travelling at the speed of $1.0 \; km/s$
$(b)$ a ball of mass $0.060 \; kg$ moving at a speed of $1.0 \; m/s$,and
$(c)$ a dust particle of mass $1.0 \times 10^{-9} \; kg$ drifting with a speed of $2.2 \; m/s$?

If a change in the momentum of an electron by $P_o$ results in a change of $0.25\%$ in its de-Broglie wavelength,then the initial momentum is:

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^{n}$,then the value of $n$ is: