Problem 87 - Master Resource Book in JEE Main Physics

Question

A metallic surface is irradiated by a monochromatic light of frequency \(v_{1}\) and stopping potential is found to be \(V_{1}\). If the light of frequency \(v_{2}\) irradiates the surface, the stopping potential will be (a) \(V_{1}+\frac{h}{e}\left(v_{1}+v_{2}\right)\) (b) \(V_{1}+\frac{h}{e}\left(v_{2}-v_{1}\right)\) (c) \(v_{1}+\frac{e}{h}\left(v_{2}-v_{1}\right)\) (d) \(V_{1}+\frac{h}{e}\left(v_{1}-v_{2}\right)\) (e) \(V_{1}-\frac{\mathrm{e}}{h}\left(v_{2}-v_{1}\right)\)

Step-by-Step Solution

Answer

The stopping potential will be \( V_1 + \frac{h}{e}(v_2 - v_1) \), which is option (b).

Step 1: Understand the Photoelectric Effect Equation

The photoelectric effect equation is given by \( eV = hu - \phi \), where \(e\) is the charge of an electron, \(V\) is the stopping potential, \(h\) is Planck's constant, \(u\) is the frequency of the incoming light, and \(\phi\) is the work function of the metal.

Step 2: Write the Equation for Frequency \(v_{1}\)

For frequency \(v_{1}\), the stopping potential is given as \(V_1\). Thus, the equation becomes \( eV_1 = hv_1 - \phi \).

Step 3: Write the Equation for Frequency \(v_{2}\)

For the new frequency \(v_{2}\), the stopping potential \(V_2\) needs to be calculated. The equation is \( eV_2 = hv_2 - \phi \).

Step 4: Solve for Work Function \(\phi\)

From Step 2, isolate \(\phi\): \( \phi = hv_1 - eV_1 \).

Step 5: Substitute \(\phi\) in the \(v_{2}\) Equation

Substitute \(\phi = hv_1 - eV_1\) into the equation from Step 3: \( eV_2 = hv_2 - (hv_1 - eV_1) \).

Step 6: Simplify the Equation for \(V_2\)

Simplify the expression: \( eV_2 = hv_2 - hv_1 + eV_1 \), which simplifies further to \( eV_2 = eV_1 + h(v_2 - v_1) \).

Step 7: Solve for Stopping Potential \(V_2\)

Isolate \(V_2\): \( V_2 = V_1 + \frac{h}{e}(v_2 - v_1) \). This matches option (b) in the given choices.

Key Concepts

Stopping Potential

The stopping potential, symbolized as \(V\), is crucial in the context of the photoelectric effect. It represents the voltage needed to stop the current in a circuit of photoelectrons ejected from a metal surface when light is shone on it. Think of it as a counterbalance to the energy of the ejected electrons. Without it, electrons would continue to flow as current due to the light’s energy.

This concept is central to experiments involving the photoelectric effect. When monochromatic light of a certain frequency illuminates a metal, electrons are emitted. The stopping potential is the measure of potential required to stop these electrons from reaching the other side in a circuit, hence stopping the electron flow.

The equation relating stopping potential to frequency is \(eV = hu - \phi\). Here, \(e\) is the charge of an electron, \(h\) is Planck's constant, \(u\) is the frequency of the incoming light, and \(\phi\) is the work function of the metal. In simpler terms, stopping potential helps balance out the light energy strong enough to dislodge electrons from the metal surface.

Work Function

The work function, denoted by \(\phi\), is another vital concept when studying the photoelectric effect. It represents the minimum energy required to remove an electron from a solid's surface. Imagine it as an energy barrier; if the incoming photon doesn't have enough energy to overcome this barrier, the electron will not be released.

This concept is intrinsic to different materials because each metal will have its unique work function value. It’s dependent on how strongly electrons are bound to the atomic lattice of that metal. If the photon energy (from the incident light) is less than the work function, no electrons will be ejected. On the other hand, if it is more significant, the excess energy is transferred to the electrons as kinetic energy, which affects their velocity.

The equation reflecting this is \(eV = hu - \phi\). By knowing either the stopping potential or the photon's frequency (\(u\)), we can determine the work function of the metal.

Planck's Constant

Planck's constant \((h)\) is a fundamental constant that underpins many quantum mechanics phenomena, including the photoelectric effect. It expresses the proportionality between the energy and frequency of a photon. In essence, it sets the scale of quantum effects – kind of like a ruler in the quantum world.

Its value is approximately \(6.626 \times 10^{-34} \text{Js}\), indicating how minute the quantum world is compared to our everyday macro world. In the photoelectric effect, it is used in the equation \(eV = hu - \phi\), connecting the light's frequency \(u\) with the energy \(eV = hu\).

Planck’s constant reveals that energy comes in discrete packets (quanta), explaining why electrons only get ejected if the photon's energy is above a certain threshold. This constant paved the way for the quantum theory, highlighting the dual particle-wave nature of light—essentially forming the basis of our understanding of light and matter at atomic and subatomic levels.