An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

$A$ nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ (where $M' < M$). The ratio of the de Broglie wavelengths of the two parts will be:

The de-Broglie wavelength of an electron with kinetic energy of $320 eV$ is (Take $h = 6.0 \times 10^{-34} \text{ SI unit}$, mass of electron $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, charge of an electron $e = 1.6 \times 10^{-19} \text{ C}$). (in $pm$)

If the de-Broglie wavelength at a temperature of $27^oC$ is $\lambda$,what will be the de-Broglie wavelength at $927^oC$?

An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation, determine the de-Broglie wavelength of the electron:

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor