Physics Circle home Student Login

Home›Chapters›Dual Nature of Radiation and Matter PYQs›Photoelectric Effect: Einstein's Equation›Question 125

JEE Main 2020 (9 Jan, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2020 (9 Jan, Shift 1)

Q125JEE Main 2020 · 9 JAN · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation

Radiation, with wavelength 6561 A˚\text{\AA} 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:

Show answer and solution

Answer: (c)

  1. Photon energy =124006561=1.89= \frac{12400}{6561} = 1.89 eV.
  2. K=eB2r22mK = \frac{eB^2r^2}{2m} eV =1.6×10−19×(3×10−4)2(10−2)22×9.1×10−31≈0.79= \frac{1.6 \times 10^{-19} \times (3 \times 10^{-4})^2 (10^{-2})^2}{2 \times 9.1 \times 10^{-31}} \approx 0.79 eV
  3. ϕ=1.89−0.79=1.10\phi = 1.89 - 0.79 = 1.10 eV
  4. So the answer is (c) 1.1 eV.
Report a mistake

More questions on Photoelectric Effect: Einstein's Equation