3→2⇒1.89eV
5×10−4Tr=7mm
r=qBmv⇒mv=qrB
⇒E=2mP2=2m(qRB)2
=2×9.1×10−31Joule(1.6×10−19×7×10−3×5×10−4)2
=18.2×10−31×1.6×10−193136×10−52eV
=1.077eV
We know work function = energy incident − (KE)electron
ϕ=1.89−1.077=0.813eV
JEE Main 2021 — Physics Modern Physics
The radiation corresponding to 3→2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5×10−4T. Assume that the radius of the largest circular path followed by these electrons is 7mm, the work function of the metal is:
(Mass of electron =9.1×10−31kg)
Held on 20 Jul 2021 · Verified 6 Jul 2026.
1.36eV
1.88eV
0.16eV
0.82eV
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