Question

In a photoelectric experiment, keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased, then the

• A. photoelectric current decreases
• B. kinetic energy of emitted photoelectrons decreases
• C. photoelectric current increases
• D. kinetic energy of emitted photoelectrons increases

Hint:

Always separate the effects of frequency and intensity in your mind: Frequency controls energy (how fast the individual electrons fly out), whereas Intensity controls quantity (how many electrons fly out per second). Since current is a measure of quantity per second, it tracks intensity directly.

Answer 1

The Correct Option is C

Step 1: Separate the two roles of light.
Frequency controls each photon's energy (hence electron kinetic energy), while intensity controls how many photons arrive per second (hence the current). Here frequency is fixed, so focus on intensity's effect on photon count.
Step 2: Kinetic energy stays put.
Einstein's relation $K_{\max} = hf - \phi$ depends only on frequency and work function. With $f$ fixed, $K_{\max}$ is unchanged, ruling out the kinetic-energy options.
Step 3: Intensity means more photons.
Higher intensity at the same frequency simply delivers more photons per second to the surface.
Step 4: One photon, one electron.
Each absorbed photon can eject one electron, so more photons per second means more electrons ejected per second.
Step 5: Link emission rate to current.
Photoelectric current is the rate of charge flow, $I = \dfrac{\Delta q}{\Delta t}$, so a higher emission rate gives a larger current.
Step 6: Conclude.
Increasing intensity raises the photoelectric current. \[ \boxed{\text{Photoelectric current increases}} \]