Problem 18 - Master Resource Book in JEE Main Physics

Question

Given that a photon of light of wavelength \(10,000 \AA\) has an energy equal to \(1.23 \mathrm{eV}\). When light of wavelength \(5000 \AA\) And intensity \(I_{0}\) falls on a photoelectric cell, the surface current is \(0.40 \times 10^{-6} \mathrm{~A}\) and the stopping potential is \(1.36 \mathrm{~V}\), then the work function is (a) \(0.43 \mathrm{eV}\) (b) \(0.55 \mathrm{eV}\) (c) \(1.10 \mathrm{eV}\) (d) \(1.53 \mathrm{eV}\)

Step-by-Step Solution

Answer

The work function is (c) 1.10 eV.

Step 1: Calculate Energy of Photon with Wavelength 5000 Å

The energy of a photon is given by the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( (4.14 \times 10^{-15} eV \cdot s) \), \( c \) is the speed of light \( (3 \times 10^8 m/s) \), and \( \lambda \) is the wavelength in meters. First, convert \( 5000 \AA \) to meters: \( 5000 \AA = 5000 \times 10^{-10} m = 5 \times 10^{-7} m \). Then plug in the values, \( E = \frac{4.14 \times 10^{-15} \times 3 \times 10^8}{5 \times 10^{-7}} \), resulting in \( E \approx 2.48 \mathrm{eV} \).

Step 2: Use Photoelectric Equation

The photoelectric equation is \( E = W + eV_0 \), where \( E \) is the energy of the incident photon, \( W \) is the work function, and \( V_0 \) is the stopping potential. We calculated \( E \approx 2.48 \mathrm{eV} \). The stopping potential \( V_0 \) is given as \(1.36 \mathrm{~V}\). Therefore, substitute these values into the equation: \( 2.48 \mathrm{eV} = W + 1.36 \mathrm{V} \).

Step 3: Solve for Work Function

Rearrange the photoelectric equation to find the work function \( W \): \( W = E - eV_0 \). Substitute the known values: \( W = 2.48 \mathrm{eV} - 1.36 \mathrm{eV} \). Calculating this gives \( W = 1.12 \mathrm{eV} \). Simplifying the options closest to \( 1.12 \mathrm{eV} \) reveals \( 1.10 \mathrm{eV} \) is the nearest value.

Key Concepts

Photon Energy

When discussing the photoelectric effect, photon energy is a key concept. A photon is a particle of light that carries energy, which can be quantified using the equation: \[E = \frac{hc}{\lambda}\]

• E stands for the energy of the photon.
• h is Planck’s constant, approximately \(4.14 \times 10^{-15} \text{eV} \cdot \text{s}\).
• c is the speed of light, roughly \(3 \times 10^8 \text{m/s}\).
• \(\lambda\) signifies the wavelength of the photon.

To find the energy, first ensure that all units are compatible. For instance, wavelengths should be converted to meters for calculations. In the case of a photon with a wavelength of \(5000 \, \text{Å}\), it converts to \(5 \times 10^{-7} \text{m}\). Using the energy formula, you would calculate its energy to be approximately \(2.48 \text{eV}\). This energy is critical in determining how a photon interacts with matter, specifically regarding the photoelectric effect.

Work Function

The work function is the minimum energy required to remove an electron from the surface of a material. It's a unique property of each material, based on its atomic structure. In the photoelectric equation:\[E = W + eV_0\]

• E is the energy of the incoming photon.
• W is the work function.
• eV_0 represents the energy associated with the stopping potential.

The work function determines if the material can emit electrons when exposed to light of a certain frequency. If the photon energy exceeds the work function, electrons are emitted, creating a current. In our exercise, by knowing the photon's energy (\(2.48 \, \text{eV}\)) and using the given stopping potential, we can isolate and solve for the work function. We found it to be \(1.10 \, \text{eV}\), meaning that this is the energy needed to overcome the electron bond on the surface of the material used in the experiment.

Stopping Potential

The stopping potential, denoted as \(V_0\), is an essential part of the photoelectric effect experiment. It represents the minimum reverse voltage needed to stop the flow of photoelectrons, thereby preventing them from reaching the other side in a photoelectric cell.In the photoelectric equation:\[E = W + eV_0\]

• E is the photon's energy.
• W is the work function.
• eV_0 is the energy contributed by the stopping potential.

The stopping potential acts as a counterforce to the kinetic energy of the emitted electrons. To calculate it, one measures the potential needed to reduce the current to zero in the photoelectric experiment. In our problem, the stopping potential \(V_0\) was \(1.36 \, \text{V}\), signifying the voltage necessary to stop the movement of electrons and measure the photoelectric current accurately. Understanding the stopping potential allows for the exploration of electron behavior under the influence of light, providing insights into the nature of light particles and their interactions.