KEmax=λhc−Ψ
⇒KEmax=5001240−2.28
⇒KEmax=2.48−2.28=0.2eV
λmin=2m(KE)maxh=2×9×10−31×0.2×1.6×10−19320×10−34
λmin=925×10−9=2.80×10−9m
So, λ≥2.8×10−9m.
Light of wavelength 500nm is incident on a metal with work function 2.28eV. The de Broglie wavelength of the emitted electron is:
Held on 30 Apr 2015 · Verified 9 Jul 2026.
<2.8×10−9m
≥2.8×10−9m
≤2.8×10−12m
<2.8×10−10m
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.