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

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ $(E_0 = \text{constant} > 0)$, with an initial velocity $\vec{V} = v_0\hat{i}$ $(v_0 = \text{constant} > 0)$. If $\lambda_0 = \frac{h}{mv_0}$, its de Broglie wavelength at time $t$ is . . . . . . .

According to the de-Broglie hypothesis,the wavelength associated with a moving electron of mass $m$ is $\lambda_e$. Using the mass-energy relation and Planck's quantum theory,the wavelength associated with a photon is $\lambda_p$. If the energy $(E)$ of the electron and the photon is the same,then the relation between $\lambda_e$ and $\lambda_p$ is:

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

De Broglie waves are associated with moving particles. These particles can be .......

If the kinetic energy of a particle is increased to $16$ times its previous value,the percentage change in the de-Broglie wavelength of the particle is