JEE MainPhysicsDual Nature of Matter
Light of wavelength 310 nm is incident on a metal surface whose work function is 2.5 eV . The stopping potential required to halt the fastest emitted photoelectrons is : (Take hc = 1240 eV nm )
Options
- A6.5 V
- B1.5 V
- C4.0 V
- D2.5 V
Correct answer
B. 1.5 V
Step-by-step solution
The energy of the incident photons is given by: E = hc E = 1240 eV nm 310 nm = 4.0 eV According to Einstein's photoelectric equation, the maximum kinetic energy ( K_ ) of the emitted photoelectrons is: K_ = E - ₀ K_ = 4.0 eV - 2.5 eV = 1.5 eV The stopping potential V_s is related to the maximum kinetic energy by K_ = eV_s . Therefore, eV_s = 1.5 eV V_s = 1.5 V Answer: 1.5 V