Question

A photoelectron emitted when a light of wavelength 2480 \(\mathring{A}\) falls on a metal, enters a uniform magnetic field of \( \frac{1}{4} \times 10^{-5} \) T perpendicular to it and moves in a circular path of maximum radius 1 m. The work function of the metal is nearly:

• A. \( 2.45 \) eV
• B. \( 3.45 \) eV
• C. \( 4.45 \) eV
• D. \( 1.45 \) eV

Hint:

- Use \( E = \frac{hc}{\lambda} \) to find photon energy. - Apply the magnetic force formula to determine the kinetic energy of the emitted electron. - Use Einstein’s photoelectric equation \( E = \phi + KE \) to determine the work function.

Answer 1

The Correct Option is C

Step 1: Energy of the Incident Photon The energy of a photon is given by: \[ E = \frac{hc}{\lambda} \] where: \( h = 6.626 \times 10^{-34} \) J.s (Planck's constant) \( c = 3.0 \times 10^8 \) m/s (Speed of light) \( \lambda = 2480 \mathring{A} = 2480 \times 10^{-10} \) m Substituting the values: \[ E = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^8)}{2480 \times 10^{-10}} \] \[ E = 8.03 \times 10^{-19} \text{ J} \] Converting to eV (\( 1 eV = 1.6 \times 10^{-19} \) J): \[ E = \frac{8.03 \times 10^{-19}}{1.6 \times 10^{-19}} = 5.02 \text{ eV} \] Step 2: Kinetic Energy of the Electron The maximum kinetic energy of the emitted photoelectron is given by: \[ KE_{\max} = \frac{e^2 B^2 R^2}{2m} \] where: \( e = 1.6 \times 10^{-19} \) C (Charge of an electron) \( B = \frac{1}{4} \times 10^{-5} \) T \( R = 1 \) m \( m = 9.1 \times 10^{-31} \) kg (Mass of an electron) Substituting: \[ KE_{\max} = \frac{(1.6 \times 10^{-19})^2 (1/4 \times 10^{-5})^2 (1)^2}{2 \times 9.1 \times 10^{-31}} \] \[ KE_{\max} \approx 0.57 \text{ eV} \] Step 3: Work Function Calculation Using the photoelectric equation: \[ \phi = E - KE_{\max} \] \[ \phi = 5.02 - 0.57 = 4.45 \text{ eV} \] Thus, the work function of the metal is \( 4.45 \) eV.