An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is

$A$ ball of mass $0.12 \ kg$ is moving with a speed $20 \ m \ s^{-1}$. Then its de Broglie wavelength is . . . . . . . ( $h = 6.63 \times 10^{-34} \ J \ s$ )

$A$ particle has a de Broglie wavelength $\lambda$. If its velocity and energy are both doubled,what will be the new de Broglie wavelength?

The de Broglie wavelength of an electron in a metal at $27^{\circ}C$ is compared with the given mean distance between two electrons in the metal,which is $2 \times 10^{-10} \ m$. The ratio of the mean distance to the de Broglie wavelength is approximately:

The electron microscope is based on the principle of

An electron microscope uses electrons accelerated by a voltage of $50\; kV$. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture,etc.) are taken to be roughly the same,how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?