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

For the same objective,find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000\,\mathring{A}$ and electrons accelerated through $100\,V$ used as the illuminating substance.

An electron beam used in an electron microscope,when accelerated by a voltage of $20\,kV$,has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to $40\,kV$,then the de-Broglie wavelength associated with the electron beam would be:

If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is

$A$ proton,an electron,and an alpha particle have the same kinetic energies. Their de-Broglie wavelengths will be compared as:

An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When a proton of mass $M$ is accelerated through a potential difference of $9V$,the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage).