In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5\times {10}^{-12}\text{m}$, the minimum electron energy required is close to

Q. We wish to see inside the atom. Assuming the atom to have a diameter of $100\text{pm}$, this means that one must be able to resolve a width of say $10\text{pm}$. If an electron-micro scope is used, the minimum electron energy required is about. (Assume the wavelength of light used in an electron microscope is nearly equal to the resolving power of the electron microscope.)

Q. We wish to seen inside an atom. Assuming the atom to have a diameter of $100pm$,this means that one must be able to resolve a width of say $10pm$. If an electron microscope is used, the minimum electron energy required is about:

Q. We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used the energy required should be

Q. In an electron microscope, electrons accelerated by a voltage of $50\text{kV}$ are used. In an optical microscope of the same aperture, yellow light of wavelength $5900\mathring{\text{A}}$ is used. The resolving power of the electron microscope will be $n$ times higher than that of the optical microscope, where $n$ is of the order of

Q. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc., to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which used yellow light?