Q185 / 200 · MHT CET - Medical 2026 (22 April Shift 2) · Physics
If the photoelectric work function of a metal is '$\phi$' and the frequency of incident light is '$\nu$', there is no emission of photoelectrons if
- Correct answerA$\nu < \dfrac{\phi}{h}$
- B$\nu = \dfrac{\phi}{h}$
- C$\nu > \dfrac{\phi}{h}$
- D$\nu \geq \dfrac{\phi}{h}$
Solution
According to Einstein's photoelectric equation, the energy of the incident photon is given by $E = h\nu$.
For photoelectric emission to take place, the energy of the incident photon must be greater than or equal to the work function of the metal, i.e., $h\nu \geq \phi$, which implies $\nu \geq \dfrac{\phi}{h}$.
Therefore, there will be no emission of photoelectrons if the energy of the incident photon is less than the work function, which means $h\nu < \phi$, or $\nu < \dfrac{\phi}{h}$.
Answer: $\nu < \dfrac{\phi}{h}$
For photoelectric emission to take place, the energy of the incident photon must be greater than or equal to the work function of the metal, i.e., $h\nu \geq \phi$, which implies $\nu \geq \dfrac{\phi}{h}$.
Therefore, there will be no emission of photoelectrons if the energy of the incident photon is less than the work function, which means $h\nu < \phi$, or $\nu < \dfrac{\phi}{h}$.
Answer: $\nu < \dfrac{\phi}{h}$
Hence, the correct answer is option: 'A'.