In a photoelectric experiment, increasing the intensity of incident light :

increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

increases the number of photons incident and also increases the K.E. of the ejected electrons

increases the frequency of photons incident and increases the K.E. of the ejected electrons

increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged

Dual Nature of Radiation PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q51 with Solution

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Correct answer: D

Photoelectric equation

For photoelectric emission, the maximum kinetic energy of emitted electrons is given by Einstein’s photoelectric equation: $K_{\max} = h\nu - \phi$ where:
- $h$ = Planck’s constant
- $\nu$ = frequency of incident light
- $\phi$ = work function of the metal
Effect of increasing intensity

Intensity of light is related to the number of photons incident per unit time (for a given frequency).

So, when intensity increases without changing frequency:
- the number of photons incident increases,
- hence the number of emitted electrons may increase,
- but the energy of each photon remains the same because photon energy is: $E = h\nu$
Effect on kinetic energy of photoelectrons

Since the frequency $\nu$ remains unchanged, from $K_{\max} = h\nu - \phi$ the maximum kinetic energy of ejected electrons also remains unchanged.
Evaluate options
- A: increases number of photons and also increases K.E. of electrons → Incorrect
- B: increases frequency and increases K.E. → Incorrect
- C: increases frequency and K.E. unchanged → Incorrect
- D: increases number of photons and K.E. unchanged → Correct
Final answer

Therefore, the correct option is: $\boxed{\text{D}}$