Write a note on the electron microscope.
(N/A) An electron microscope uses a beam of electrons instead of visible light to image specimens.
The electron beam is focused using precisely designed electric and magnetic fields,which act as lenses.
The resolution of an electron microscope is limited by the wave nature of electrons. According to de Broglie's hypothesis,the wavelength of an electron is much smaller than that of visible light,typically less than $1 \; \mathring{A}$.
Due to this extremely short wavelength,electron microscopes can achieve a resolution of approximately $0.6 \; \mathring{A}$.
Because of this high resolution,electron microscopes are used to observe and resolve individual atoms and molecules.
The electron beam is focused using precisely designed electric and magnetic fields,which act as lenses.
The resolution of an electron microscope is limited by the wave nature of electrons. According to de Broglie's hypothesis,the wavelength of an electron is much smaller than that of visible light,typically less than $1 \; \mathring{A}$.
Due to this extremely short wavelength,electron microscopes can achieve a resolution of approximately $0.6 \; \mathring{A}$.
Because of this high resolution,electron microscopes are used to observe and resolve individual atoms and molecules.
Explore More
Similar Questions
The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)
DifficultTS EAMCET 2012
View SolutionThe de-Broglie wavelength of an electron having kinetic energy $100 eV$ is, $[$ Use $h=4.14 \times 10^{-15} eVs$, mass of electron $= \frac{0.5 \times 10^6}{c^2} eV/c^2$, $1 pm = 10^{-12} m$ $]$ (in $pm$)
$A$ charged particle is accelerated from rest through a certain potential difference. The de-Broglie wavelength is $\lambda_1$ when it is accelerated through $V_1$ and is $\lambda_2$ when accelerated through $V_2$. The ratio $\lambda_1 / \lambda_2$ is
$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to:
Difficult
View SolutionThe kinetic energies of an electron,$\alpha$-particle,and a proton are given as $4K, 2K$,and $K$ respectively. The de-Broglie wavelengths associated with the electron $(\lambda_e)$,$\alpha$-particle $(\lambda_\alpha)$,and the proton $(\lambda_p)$ are related as follows:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo