KEmax=E−ϕ
=λ(inÅ)12400−ϕ (in eV)
∴r=eB2mKE
KEmax=2mr2e2B2 (in J)
=2mr2eB2 (in eV)
∴ϕ=655612400−2mr2eB2
=1.1eV
JEE Main 2020 — Physics Electromagnetism
Radiation, with wavelength 6561A˚ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3×10−4T . If the radius of the largest circular path followed by the electrons is 10mm , the work function of the metal is close to:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
1.6eV
0.8eV
1.1eV
1.8eV
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