Problem 19 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

If \(\lambda_{0}\) and \(\lambda\) be threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is: (a) \(\sqrt{\frac{2 h}{m}\left(\lambda_{0}-\lambda\right)}\) (b) \(\sqrt{\frac{2 h c}{m}\left(\lambda_{0}-\lambda\right)}\) (c) \(\sqrt{\frac{2 h c}{m}\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{0}}\right)}\) (d) \(\sqrt{\frac{2 h}{m}\left(\frac{1}{\lambda_{0}}-\frac{1}{\lambda}\right)}\)

Step-by-Step Solution

Answer

The answer is (d): \(\sqrt{\frac{2 h c}{m}\left(\frac{1}{\lambda_{0}}-\frac{1}{\lambda}\right)}\).

Step 1: Understand the Photoelectric Equation

The photoelectric effect relates the energy of an incident photon to the kinetic energy of an emitted photoelectron. The energy needed to remove an electron from the metal is given by \(\phi = \frac{hc}{\lambda_0}\), where \(\lambda_0\) is the threshold wavelength.

Step 2: Determine Kinetic Energy Using Wavelengths

Using the relation \(E_{photon} - E_{threshold} = KE_{electron}\), we have: \[KE = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}\].

Step 3: Simplify the Kinetic Energy Equation

Rewriting, we get \[KE = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)\].

Step 4: Calculate Velocity from Kinetic Energy

The kinetic energy \(KE\) is also related to velocity \(v\) by \(KE = \frac{1}{2}mv^2\). Equating and solving for \(v\), we have \[\frac{1}{2}mv^2 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)\].

Step 5: Solve for Velocity \(v\)

Rearrange to find \(v\): \[v = \sqrt{\frac{2hc}{m}\left(\frac{1}{\lambda_0} - \frac{1}{\lambda}\right)}\]. This matches option (d).

Key Concepts

Threshold Wavelength

When we discuss the photoelectric effect, one of the primary concepts is the threshold wavelength, \( \lambda_0 \). This is the particular wavelength of light that defines the minimum energy required to eject an electron from a metal surface. Each type of metal has its own unique threshold wavelength, tied directly to its work function \( \phi \).
The work function is the energy needed to liberate an electron from the surface of the metal.

• The energy of a photon is inversely proportional to its wavelength, given by the formula: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light.
• At the threshold wavelength, the energy provided by the incident photons precisely equals the work function of the metal. Thus, no electron ejection occurs for photons with a wavelength greater than \( \lambda_0 \).
• A photon with wavelength \( \lambda_0 \) provides exactly enough energy to overcome the binding energy but no excess energy is available to give the ejected electron any kinetic energy.

Understanding the threshold wavelength allows us to predict whether an electron will be ejected when light hits a metal, making it a crucial component in discussing and experimenting with the photoelectric effect.

Kinetic Energy of Photoelectrons

The kinetic energy of photoelectrons is the energy that an electron possesses as it moves away after being ejected from a metal surface. This energy can be calculated as the difference between the energy of the incident photon and the work function of the metal.

• Using Einstein's photoelectric equation, the kinetic energy \( KE \) is given by: \( KE = E_{photon} - E_{threshold} \).
• Substituting the expressions for the energies, we have: \[ KE = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \].
• Simplifying further leads to: \[ KE = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \].

This relation emphasizes that kinetic energy is directly influenced by the difference in the inverse of the wavelengths. It shows that for wavelengths shorter than the threshold, the photon supplies more energy, which contributes to the kinetic energy of the photoelectrons. This equation is fundamental for understanding how changes in the wavelength of incident light affect the speed of ejected electrons.

Equation for Velocity of Ejected Electrons

Determining the velocity of ejected electrons gives insight into how quickly an electron is expelled from a metal surface during the photoelectric effect. The velocity \( v \) is derived from the kinetic energy formula used earlier.

• We relate kinetic energy and velocity using the equation: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron.
• By combining this with the earlier kinetic energy formula, \( \frac{1}{2}mv^2 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \), we solve for velocity.
• Rearranging gives: \[ v = \sqrt{\frac{2hc}{m} \left( \frac{1}{\lambda_0} - \frac{1}{\lambda} \right)} \] which highlights the dependency of the velocity on the inverses of the wavelengths.

This equation serves as a key to understanding how the energy transferred from light affects the speed at which electrons can leave the metal. When the wavelength of the incoming photon decreases (increasing the energy), the velocity of the ejected electrons increases, demonstrating the interplay between light and matter in the context of the photoelectric effect.