JEE MainPhysicsDual Nature of Matter
The work functions of metal A and metal B are 2.0 eV and 3.1 eV , respectively. When monochromatic light of wavelength is incident on both the metals, photoelectric emission is observed from metal A but not from metal B. Which of the following could be the value of ? (Take hc = 1240 eV nm )
Options
- A310 nm
- B650 nm
- C550 nm
- D380 nm
Correct answer
C. 550 nm
Step-by-step solution
For photoelectric emission to occur, the energy of the incident photon must be greater than the work function of the metal ( E > ). Given that emission occurs for metal A but not for metal B, the incident photon energy E must satisfy: _A 2.0 eV The energy of a photon is given by E = hc = 1240 eV , where is in nm. Checking the given options: For = 310 nm , E = 1240 310 = 4.0 eV (Emission from both) For = 650 nm , E = 1240 650 1.91 eV (Emission from neither) For = 550 nm , E = 1240 550 2.25 eV (Emission from A only)