Question

A metal surface is illuminated by a radiation of wavelength 4500 Å. The rejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately:

• A. 1.36 eV
• B. 1.69 eV
• C. 2.78 eV
• D. 2.23 eV

Answer 1

The Correct Option is A

To solve this problem, we will use the principles of photoelectric effect and motion of charged particles in a magnetic field.

1. The photon energy incident on the metal is given by the formula: E = \frac{hc}{\lambda} where:
   • h is Planck's constant, 6.626 \times 10^{-34} \text{ J s}
   • c is the speed of light, 3 \times 10^{8} \text{ m/s}
   • \lambda is the wavelength, 4500 \text{ \AA} = 4500 \times 10^{-10} \text{ m}
   Substituting these values, we find: E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4500 \times 10^{-10}} E \approx 4.407 \times 10^{-19} \text{ J}
2. Convert this energy from joules to electronvolts (1 eV = 1.602 \times 10^{-19} \text{ J}): E \approx \frac{4.407 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.75 \text{ eV}
3. When the emitted electron enters the magnetic field, it follows a circular path. The force due to the magnetic field provides the necessary centripetal force: F = qvB = \frac{mv^2}{r} where:
   • q is the charge of the electron, 1.6 \times 10^{-19} \text{ C}
   • v is the velocity of the electron
   • B is the magnetic field, 2 \times 10^{-3} \text{ T}
   • r is the radius of the path, 2 \times 10^{-3} \text{ m}
   Solving for v gives: v = \frac{qBr}{m} Substituting, v = \frac{1.6 \times 10^{-19} \times 2 \times 10^{-3} \times 2 \times 10^{-3}}{9.11 \times 10^{-31}} v \approx 7 \times 10^{5} \text{ m/s}
4. The kinetic energy of the emitted electron is given by: KE = \frac{1}{2}mv^2 \approx \frac{1}{2} \times 9.11 \times 10^{-31} \times (7 \times 10^{5})^2 KE \approx 2.24 \times 10^{-19} \text{ J} Converting this to electronvolts: KE = \frac{2.24 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.4 \text{ eV}
5. Finally, apply the photoelectric effect equation: E = KE + \phi\ ( \text{work function, } \phi) 2.75 = 1.4 + \phi \Rightarrow \phi \approx 1.35 \text{ eV} Thus, the work function of the metal is approximately 1.36 eV, which matches the given option.

Conclusion: Therefore, the correct answer is 1.36 eV.