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 \(7.5\times10^{-12}\) m, the minimum electron energy required is close to -
Show Hint
- 25 keV
- 500 keV
- 100 keV
- 1 keV
The Correct Option is A
Solution and Explanation
\(K = \frac{(6.63\times10^{-34})^2}{2(9.1\times10^{-31})(7.5\times10^{-12})^2} = \frac{4.40\times10^{-67}}{1.02\times10^{-52}} \approx 4.3\times10^{-15}\) J.
In eV: \(\frac{4.3\times10^{-15}}{1.6\times10^{-19}} \approx 2.7\times10^{4}\) eV \(\approx 27\) keV, closest to 25 keV.
Answer: (1) 25 keV
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
Top JEE Main de broglie hypothesis Questions
- Determine ratio of de broglie wavelength of \(α- particle\) and proton of same kinetic energy is
- De Broglie wavelength of proton = \(λ\) and that of an a particle \(2λ\). The ratio of velocity of proton to that of \(\alpha\) particle is
The de Broglie wavelengths of a proton and an α particle are \( \lambda \) and \( 2\lambda \) respectively. The ratio of the velocities of proton and α particle will be:
The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:
- The de-Broglie wavelength associated with a particle of mass \( m \) and energy \( E \) is \[\lambda = \frac{h}{\sqrt{2mE}}.\]The dimensional formula for Planck's constant is:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,