The electron microscope is based on the principle of

An electron (mass $m$) with an initial velocity $v = v_0 \hat{i}$ is in an electric field $E = E_0 \hat{j}$. If $\lambda_0 = h/mv_0$,its de Broglie wavelength at time $t$ is given by

When an electron is accelerated through a potential $V$,the de-Broglie wavelength associated with it is $4 \lambda$. When the accelerating potential is increased to $4V$,its wavelength will be

The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0, & 0 \le x \le 1 \\ 0, & x > 1 \end{cases}$. Let $\lambda_1$ and $\lambda_2$ be the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2E_0$,find $(\lambda_1/\lambda_2)^2$.

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

The de Broglie wavelength of an oxygen molecule at $27^{\circ} C$ is $x \times 10^{-12} \ m$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34} \ J \cdot s$, Boltzmann constant $= 1.38 \times 10^{-23} \ J/K$, mass of oxygen molecule $= 5.31 \times 10^{-26} \ kg$).