Verified Solution • 2025–26 Session moderate

QJEEM20Ph4728: JEE Main 2020 Physics - Class 12 JEEM20Ph4728 Solution: Radiation, with wavelength 6561 falls on a metal surface to…

Radiation, with wavelength 6561 Ao\mathop A\limits^o falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :

Multiple Choice Options

  • A 1.8eV
  • B 0.8eV
  • C 1.1eV
  • D 1.6eV

Step-by-Step Verified Solution

Let the work function be ϕ\phi .

∴\therefore KEmax = hcλ−ϕ{{hc} \over \lambda } - \phi

We know r = mvqB{{mv} \over {qB}}

and p = mv = rqB

∴\therefore KEmax = p22m{{{p^2}} \over {2m}} = q2r2B22m{{{q^2}{r^2}{B^2}} \over {2m}}

= (1.6×10−19)2(10×10−3)2(3×10−4)22×9×10−31×1.6×10−19{{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}{{\left( {10 \times {{10}^{ - 3}}} \right)}^2}{{\left( {3 \times {{10}^{ - 4}}} \right)}^2}} \over {2 \times 9 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}}}}

= 0.8 eV

hcλ{{hc} \over \lambda } = 124206561{{12420} \over {6561}} = 1.9 eV

∴\therefore ϕ\phi = 1.9 - 0.8 = 1.1 eV

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