Problem 49 - Master Resource Book in JEE Main Physics

Question

When a monochromatic point source of light is at a distance of \(0.2 \mathrm{~m}\) from a photo electric cell, the cut off voltage and the saturation currents are respectively \(0.6 \mathrm{v}\) and \(18.0 \mathrm{~mA}\). If, the same source is placed \(0.6 \mathrm{~m}\) away from the photoelectric cell, then (a) the stopping potential will be \(0.2 \mathrm{~V}\) (b) the stopping potential will be \(0.6 \mathrm{~V}\) (c) the saturation current will be \(0.6 \mathrm{~mA}\) (d) the saturation current will be \(0.2 \mathrm{~mA}\)

Step-by-Step Solution

Answer

(b) the stopping potential remains 0.6V; (d) saturation current becomes approximately 2.0 mA.

Step 1: Understand the photoelectric effect

In the photoelectric effect, when light of a certain frequency shines on a metal, it can emit electrons. The stopping potential (cut-off voltage) is the minimum voltage needed to stop the emitted electrons. A key point is that the stopping potential depends only on the frequency of the light and not on its intensity.

Step 2: Assess effect of light distance change on stopping potential

The stopping potential is related to the energy of the photons, which is determined by the light's frequency. It remains constant as long as the frequency of the light does not change. Therefore, moving the light source further from the cell does not change the stopping potential. Thus, the stopping potential remains at \(0.6 \text{ V}\).

Step 3: Consider impact on saturation current

The saturation current relates directly to the number of photoelectrons emitted, which depends on the intensity of the light (the amount of photons incident on the surface). As the light source is moved further away from the photoelectric cell, the intensity reaching the cell decreases, leading to a reduced saturation current. The current reduces proportional to \(1/r^2\).

Step 4: Calculate new saturation current

Originally, the distance is \(0.2 \text{ m}\), and saturation current is \(18.0 \text{ mA}\). When the distance becomes \(0.6 \text{ m}\), the intensity reduces by factor of \((\frac{0.2}{0.6})^2\). Thus, \[ I_{new} = 18.0 \text{ mA} \times \left(\frac{0.2}{0.6}\right)^2 = 18.0 \text{ mA} \times \frac{1}{9} = 2.0 \text{ mA} \].

Key Concepts

Stopping Potential

The stopping potential is a fundamental concept in understanding the photoelectric effect. It represents the minimum voltage needed to stop the emitted electrons from a metal surface when exposed to light. This voltage is crucial because no matter how intense the light is, the stopping potential is determined solely by the frequency of the light.

In this context, you might wonder why the stopping potential does not change when the light source is moved further away. This is because the stopping potential depends on the energy of the light photons, which is related to their frequency, not their intensity. Therefore, as long as the frequency remains constant, the stopping potential will also stay unchanged. For instance, in the given exercise, even when the light source is moved from 0.2 m to 0.6 m, the stopping potential remains at 0.6 V.

Saturation Current

Saturation current is an important aspect of the photoelectric effect, as it indicates the maximum current possible when all emitted electrons are collected. This current is directly influenced by the number of photoelectrons emitted, which is proportional to the intensity of the incident light.

In the photoelectric effect, when a light source is positioned further from the photoelectric cell, the intensity of light reaching the cell decreases. This decrease in intensity leads to a reduction in the number of photoelectrons emitted, hence a lower saturation current. For instance, if you move the light source from 0.2 m to 0.6 m, the intensity reduces by the square of the distance ratio

• Original distance: 0.2 m
• New distance: 0.6 m

Therefore, the new saturation current is given by: \[I_{new} = 18.0 \text{ mA} \times \left(\frac{0.2}{0.6}\right)^2 = 2.0 \text{ mA}\]
This calculation shows how intensity affects saturation current, but doesn't impact stopping potential.

Intensity of Light

The intensity of light is a pivotal factor in determining the saturation current in the photoelectric effect. Intensity refers to the power per unit area received by the surface of the photoelectric cell. The closer a light source is to the cell, the higher the intensity and vice versa.

Understanding intensity is essential because, as the intensity increases, more photons hit the surface, increasing the number of electrons emitted. This increase in emitted electrons leads to a higher saturation current. In the exercise, moving the light source further away diminished the intensity due to the inverse square law. This law states:

• Intensity is inversely proportional to the square of the distance from the source.
• As distance triples (from 0.2 m to 0.6 m), intensity becomes nine times weaker.

Such a dramatic decrease in intensity from 18.0 mA to 2.0 mA shows the power of this inverse relationship, underlining that while intensity impacts saturation current, it does not affect the stopping potential.