The de-Broglie wavelength $\lambda$ associated with an electron having kinetic energy $E$ is given by the expression:

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 $2.2 \; m/s$?

We wish to see inside an atom. Assuming the atom to have a diameter of $100 \ pm$, this means that one must be able to resolve a width of say $10 \ pm$. If an electron microscope is used, the minimum electron energy required is about....... $KeV$.

An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

If the kinetic energy of the particle is increased by $16$ times,the percentage change in the de Broglie wavelength of the particle is ............$\%$

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