Exams › JEE Main › Physics

The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10⁻⁴ T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:

1. 1.8 eV
2. 1.1 eV
3. 0.8 eV
4. 1.6 eV

Correct answer: 1.1 eV

Solution

The work function of the metal can be determined using the energy of the emitted photoelectrons, which is derived from the difference between the photon energy from the 3 → 2 transition and the work function. Given the radius of the circular path in the magnetic field, we can calculate the kinetic energy of the electrons, leading to the conclusion that the work function is approximately 1.1 eV.

Related JEE Main Physics questions

• A star is sending out light of wavelength 5000 Å and is moving toward the Earth with a speed of 1.50 × 10⁶ m/s. What is the shift in the wavelength of the light observed on Earth?
• A beam of radiation carrying energy E is incident perpendicularly on a perfectly reflecting surface. What momentum is imparted to the surface? (C = speed of light)
• Two identical photocathodes are illuminated by light of frequencies f1 and f2. If the emitted photoelectrons have speeds v1 and v2 respectively, then
• An X-ray tube is run with a potential difference of 15 kV. What is the maximum possible speed of the electrons that hit the target?
• In a photoelectric setup, monochromatic light of wavelength λ ejects electrons, and the maximum speed of the emitted electrons is v. If the incident wavelength is changed to 5λ/4, the speed of the emitted electrons will be
• A sodium street lamp rated at 200 W produces yellow radiation of wavelength 0.6 μm. If only 25% of the electrical power is converted into light, how many yellow photons are emitted each second?

⚔️ Practice JEE Main Physics free + battle 1v1 →