From Einstein's equation,
Eincident=ϕ+kEmax
hf=0+21mv2
λhc=2mp2⇒p=λ2mhc
λd=ph=λ2mhch=2mcλh
⇒λ=(h2mc)λd2
An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If m mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then :
Held on 30 Apr 2021 · Verified 9 Jul 2026.
λ=(hc2m)λd2
λd=(h2mc)λ2
λ=(h2mc)λd2
λ=(mc2h)λd2
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
De-Broglie wavelength of an electron orbiting in the $n=2$ state of hydrogen atom is close to (Given Bohr radius $=0.052 \mathrm{~nm}$ )
Which of the following options represent the variation of photoelectric current with property of light shown on the x -axis? 
The maximum kinetic energy of photoelectrons from a surface with work function φ = 2 eV when light of λ = 4000 Å falls on it is (hc = 12400 eV·Å):
A full wave rectifier circuit with diodes $\left(D_1\right)$ and $\left(D_2\right)$ is shown in the figure. If input supply voltage $V_{\text {in }}=220 \sin (100 \pi t)$ volt, then at $t=15 \mathrm{msec}$ 
A particle of mass $m$ is moving around the origin with a constant force $F$ pulling it towards the origin. If Bohr model is used to describe its motion, the radius $r$ of the $n^{\text {th }}$ orbit and the particle's speed $v$ in the orbit depend on $n$ as
Work through every NEET UG Modern Physics PYQ, year by year.