Question
Explain the effect of intersity of light on photocurrent.
✓
Answer
→In an experimental arrangement of the photoelectric effect the collector A is maintained at a positive potential with respect to photosensitive plate C .
→The potential difference between A and C and the frequency of the incident light is kept constant.
→The intensity of incident light is changed.
→When light is incident on plate C , electrons are emitted from it and these are attracted by plate A to form a photocurrent.
→The magnitude of photocurrent obtained for different light intensities is measured with a microammeter and a graph of photocurrent versus intensity is drawn.
→The straight line of graph suggests that the photocurrent is directly proportional to the intensity of incident light.
→Now, current $I =\frac{q}{t}$
$\therefore I =\frac{n e}{t} \quad(\because q=n e)$
BDkt, $n=$ Number of electrons
$e=$ Charge of electron
$\therefore \quad I \propto \frac{n}{t} \quad(\because e$ is constant $)$
→Thus, the current (photocurrent) is directly proportional to the number of photoelectrons emitted per unit time.
→From this it can be said that the number of photoelectrons emitted per unit time (one second) is directly proportional to the intensity of incident radiation.
Photocurrent $\propto$ Intensity
Photocurrent $\propto$ number of electrons per unit time. (Number of electrons emitted per unit time) $\propto$ (Intensity)
→The potential difference between A and C and the frequency of the incident light is kept constant.
→The intensity of incident light is changed.
→When light is incident on plate C , electrons are emitted from it and these are attracted by plate A to form a photocurrent.
→The magnitude of photocurrent obtained for different light intensities is measured with a microammeter and a graph of photocurrent versus intensity is drawn.
→The straight line of graph suggests that the photocurrent is directly proportional to the intensity of incident light.
→Now, current $I =\frac{q}{t}$
$\therefore I =\frac{n e}{t} \quad(\because q=n e)$
BDkt, $n=$ Number of electrons
$e=$ Charge of electron
$\therefore \quad I \propto \frac{n}{t} \quad(\because e$ is constant $)$
→Thus, the current (photocurrent) is directly proportional to the number of photoelectrons emitted per unit time.
→From this it can be said that the number of photoelectrons emitted per unit time (one second) is directly proportional to the intensity of incident radiation.
Photocurrent $\propto$ Intensity
Photocurrent $\propto$ number of electrons per unit time. (Number of electrons emitted per unit time) $\propto$ (Intensity)
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