λ=V150⇒7.5×10−2=V150
V=7.5×7.5×10−4150=380kV
V≈25KeV
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 7.5×10−12m, the minimum electron energy required is close to:
Held on 10 Jan 2019 · Verified 6 Jul 2026.
100keV
25keV
1keV
500keV
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is $\_\_\_\_$ $\Omega$.
An atom ${ }_{3}^{8} X$ is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by:
In the hydrogen atom, the electron makes a transition from the higher orbit ($i$) to a lower orbit ($f$). The ratio of the radius of the orbits in given by $r_i : r_f = 16 : 4$. The wavelength of photon emitted due to this transition is _____ nm. (Given Rydberg constant $= 1.0973 \times 10^7$ /m)
The work function of a metal is 4.2 eV. The threshold wavelength for photoelectric emission is approximately:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The electromagnetic wave exerts pressure on the surface on which they are allowed to fall. Reason (R): There is no mass associated with the electromagnetic waves. In the light of the above statements, choose the correct answer from the options given below :
Work through every JEE Main Modern Physics PYQ, year by year.