The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

The de Broglie wavelength of an electron of kinetic energy $9 \ eV$ is (take $h=4 \times 10^{-15} \ eV \cdot s$,$c=3 \times 10^{10} \ cm/s$ and the mass $m_e$ of electron as $m_e c^2=0.5 \ MeV$)

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron 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 $(\lambda_1/\lambda_2)^2$.

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$? (Let $E$ be the initial energy of the electron).