We have, λp=pph and λe=peh
Given here, λe=λp, then pp=pe
Ratio of their energy is EphEe=hc\lambda Ppe22m=2mcpe=2mcmv=2cv
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and Pe and that of photon are Eph and Pph respectively. Which of the following is correct?
Held on 26 Jun 2022 · Verified 6 Jul 2026.
EphEe=v2c
EphEe=2cv
pphpe=v2c
pphpe=2cv
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.