Problem 18 - Master Resource Book in JEE Main Physics

Question

Radiations of two photon's energy, twice and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of photoelectrons emitted in two cases is (a) \(1: 2\) (b) \(1: 3\) (c) \(1: 4\) (d) \(1: 1\)

Step-by-Step Solution

Answer

The ratio of maximum velocities of photoelectrons emitted is \(1:3\).

Step 1: Understanding Work Function and Photon Energy

The work function of a metal, denoted as \( \phi \), is the minimum energy required to remove an electron from the metal surface. Photons incident on the metal have energies \( 2\phi \) and \( 10\phi \) respectively in the two cases.

Step 2: Applying Einstein's Photoelectric Equation

Einstein's photoelectric equation is given by \( E = K_{\text{max}} + \phi \), where \( E \) is the energy of the photon and \( K_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectron. For photon energies \( E_1 = 2\phi \) and \( E_2 = 10\phi \), substitute these into the equation for each case.

Step 3: Calculating Kinetic Energy for Each Case

For the first photon \( E_1 = 2\phi \): \( K_{1, \text{max}} = E_1 - \phi = 2\phi - \phi = \phi \).For the second photon \( E_2 = 10\phi \): \( K_{2, \text{max}} = E_2 - \phi = 10\phi - \phi = 9\phi \).

Step 4: Relating Kinetic Energy to Maximum Velocity

The maximum kinetic energy can be expressed as \( K_{\text{max}} = \frac{1}{2}mv^2 \), where \( m \) is the electron mass and \( v \) is the velocity. Thus:\( v_1 = \sqrt{\frac{2K_{1, \text{max}}}{m}} = \sqrt{\frac{2\phi}{m}} \) and\( v_2 = \sqrt{\frac{2K_{2, \text{max}}}{m}} = \sqrt{\frac{18\phi}{m}} \).

Step 5: Calculating the Ratio of Velocities

Substitute the expressions for velocities:\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{2\phi}{m}}}{\sqrt{\frac{18\phi}{m}}} = \sqrt{\frac{2}{18}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \].Thus, the ratio of maximum velocities is \( 1:3 \).

Key Concepts

Work Function

In the realm of physics, the term "work function" is pivotal when studying the interaction of light with metals. It refers to the minimum energy required to remove an electron from the surface of a metal. This energy is usually measured in electron volts (eV). The concept of work function is essential to understanding how light can cause a metal to emit electrons in a phenomenon known as the photoelectric effect.
Whenever light, particularly in the form of photons, strikes the surface of a metal, these photons must possess at least the energy equal to the work function of that metal to successfully release electrons.
To put it simply, if the photon energy is less than the work function, no electron emission occurs. If it meets or exceeds this threshold, electrons are emitted. In our problem, the energies of the incoming photons are described as multiples of the work function, specifically twice \(2\phi\) and ten times \(10\phi\).

• Twice the work function (\(2\phi\)): Here, the photon energy is only double the minimum required energy.
• Ten times the work function (\(10\phi\)): In this scenario, the photon energy significantly surpasses the minimum threshold.

This concept is crucial in predicting the computation and behavior of electrons ejected from the metal surface.

Photon Energy

Photon energy is a cornerstone of the photoelectric effect. It is the energy carried by a photon, which depends directly on its frequency, as described by the equation \( E = hu \), where \( E \) is the photon energy, \( h \) is Planck's constant, and \( u \) is the frequency of the photon.
In the context of the problem:

• The first photon has an energy of \( 2\phi \), which is enough to overcome the work function and impart some kinetic energy to the emitted electrons.
• The second photon, with \( 10\phi \), not only overcomes the work function but also provides far greater kinetic energy to the electrons.

When discussing photon energy, it’s important to remember that higher energy photons result in higher kinetic energy of emitted electrons, provided they exceed the work function.
This directly scales the potential for photoelectron emission, impacting the energy and speed of those electrons.

Einstein's Photoelectric Equation

Einstein's photoelectric equation is a fundamental relationship that describes the photoelectric effect, developed in 1905. It relates the energy of incident photons to the kinetic energy of emitted electrons with the formula: \[ E = K_{\text{max}} + \phi \]Where:

• \(E\) is the energy of the photon.
• \(K_{\text{max}}\) is the maximum kinetic energy of the emitted electron.
• \(\phi\) is the work function of the metal.

This equation underscores the principle that any photon energy exceeding the work function translates into kinetic energy of the photoelectron.
This energy transfer contributes to freeing electrons if the photon energy surpasses the work function.
For instance, substituting energies \(E_1 = 2\phi\) and \(E_2 = 10\phi\) for the two cases:

• When \(E_1 = 2\phi\), the maximum kinetic energy \(K_{1, \text{max}} = 2\phi - \phi = \phi\).
• For \(E_2 = 10\phi\), \(K_{2, \text{max}} = 10\phi - \phi = 9\phi\).

This clearly demonstrates how photon energy allocation either liberates or energizes electrons.

Kinetic Energy of Photoelectrons

The kinetic energy of photoelectrons is the energy they possess owing to their emission from the metal surface upon absorbing photon energy. Derived from Einstein's photoelectric equation, it directly depends on the incident photon energy and the metal's work function.
For photoelectrons, the kinetic energy \(K\) is further mathematically expressed as:\[ K_{\text{max}} = \frac{1}{2}mv^2 \]Here \(m\) is the mass of the electron, and \(v\) symbolizes their velocity.
In our specific problem, we find the kinetic energies of the photoelectrons for two incidents:

• For \(E_1 = 2\phi\): \(K_{1, \text{max}} = \phi\), leading to \(v_1 = \sqrt{\frac{2\phi}{m}}\).
• For \(E_2 = 10\phi\): \(K_{2, \text{max}} = 9\phi\), resulting in \(v_2 = \sqrt{\frac{18\phi}{m}}\).

These calculations illustrate that the velocities of the photoelectrons, proportional to their kinetic energies, determine how fast the electrons were propelled from the metal upon light exposure.
The resulting velocity ratio \((v_1 : v_2)\) is revealed as \(1:3\), reflecting the effect of differing photon energies on electron kinetics.