At closest approach, KE converts to PE:
KE=4πε01×rmin(2e)(Ze)
rmin=7.7×106×1.6×10−199×109×2×79×(1.6×10−19)2
=7.79×2×79×1.6×10−19×103
=7.72275.2×10−16=2.95×10−14 m
If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold =79 and 4πϵo1=9×109 in SI units)
Held on 21 Jan 2026 · Verified 6 Jul 2026.
3.85×10−16
3.85×10−14
2.95×10−16
2.95×10−14
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.