Problem 17 - Master Resource Book in JEE Main Physics

Question

What is the strength of transverse magnetic field required to bend all the photoelectrons within a circle of a radius \(50 \mathrm{~cm}\). When light of wavelength \(3800 \AA\) Ais incident on a barium emitted? (Given that work function of barium is \(2.5 \mathrm{eV} ; h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}\); \(\left.e=1.6 \times 10^{-19} \mathrm{C} ; m=9.1 \times 10^{-31} \mathrm{~kg}\right)\) (a) \(6.32 \times 10^{-4} \mathrm{~T}\) (b) \(6.32 \times 10^{-5} \mathrm{~T}\) (c) \(6.32 \times 10^{-6} \mathrm{~T}\) (d) \(6.32 \times 10^{-8} \mathrm{~T}\)

Step-by-Step Solution

Answer

(a) \(6.32 \times 10^{-4} \ \text{T}\)

Step 1: Calculate the Energy of the Incident Photon

First, we need to calculate the energy of the incident photon using the formula \( E = \frac{hc}{\lambda} \). Here, the speed of light \( c = 3 \times 10^8 \ \text{m/s} \) and the Planck's constant \( h = 6.63 \times 10^{-34} \ \text{J s} \). The wavelength \( \lambda = 3800 \ \text{Å} = 3800 \times 10^{-10} \ \text{m} \).\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3800 \times 10^{-10}} = 5.22 \times 10^{-19} \ \text{J} \]

Step 2: Convert Work Function to Joules

The work function of barium is given in electronvolts: \( 2.5 \ \text{eV} \). Convert it to joules using \( 1 \ \text{eV} = 1.6 \times 10^{-19} \ \text{J} \).\[ \Phi = 2.5 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J/eV} = 4.0 \times 10^{-19} \ \text{J} \]

Step 3: Calculate Maximum Kinetic Energy of the Photoelectrons

Using the photoelectric equation \( K_{\text{max}} = E - \Phi \) where \( E \) is the energy of the photon calculated in Step 1.\[ K_{\text{max}} = 5.22 \times 10^{-19} - 4.0 \times 10^{-19} = 1.22 \times 10^{-19} \ \text{J} \]

Step 4: Relate Kinetic Energy to Velocity of Photoelectrons

The relation between kinetic energy and velocity is given by \( K_{\text{max}} = \frac{1}{2} mv^2 \). Solve for velocity \( v \).\[ 1.22 \times 10^{-19} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2 \]\[ v^2 = \frac{2 \times 1.22 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 2.68 \times 10^{11} \]\[ v = \sqrt{2.68 \times 10^{11}} \approx 5.18 \times 10^5 \ \text{m/s} \]

Step 5: Calculate Magnetic Field Strength Required

Photoelectrons move in a circle due to the magnetic force which provides the centripetal force: \( eBv = \frac{mv^2}{r} \). Solve for \( B \).\[ B = \frac{mv}{er} = \frac{9.1 \times 10^{-31} \times 5.18 \times 10^5}{1.6 \times 10^{-19} \times 0.5} \]\[ B = 6.32 \times 10^{-4} \ \text{T} \]

Key Concepts

Transverse Magnetic Field

The concept of a transverse magnetic field is crucial when examining how charged particles such as photoelectrons move under the influence of magnetic forces. A transverse magnetic field is one that is oriented perpendicular to the direction of motion of these charged particles.
This perpendicular arrangement causes the particles to experience a force that is perpendicular to both the field and their velocity. As a result, the charged particles move in a circular path.
This circular motion is achieved because the magnetic force acts as the centripetal force needed to keep the particle in motion within a circle.

• The formula for the magnetic force is given by: \( F = eBv \), where
  \( e \) is the charge of the electron,
  \( B \) is the magnetic field strength, and
  \( v \) is the velocity of the particle.

The force acts inward toward the center of the circle, keeping the photoelectron on its path. This relationship helps us determine the required magnetic field strength to bend photoelectrons, like in this exercise, where they describe a circle of a given radius.

Photoelectron Kinetic Energy

Kinetic energy of photoelectrons is the energy acquired by them once they are ejected from a material under light exposure. The relationship between the energy of the incident light and the energy needed to release these electrons influences this kinetic energy.

According to the photoelectric effect principle, the kinetic energy (\( K \) of a photoelectron is calculated using the equation:

• \( K = E - \Phi \)
  Where:
  \( E \) is the energy of the photon (calculated as \( E = \frac{hc}{\lambda} \)), and
  \( \Phi \) is the work function of the material.

The photon energy initially comes from the light hitting the material, while the work function is the minimum energy needed to free an electron from that material.
Therefore, the kinetic energy is what remains after this required energy is used.

In our problem, we calculate the kinetic energy of barium's photoelectrons, showing how much excess energy they have to start moving, once released.

Work Function

The work function is an essential concept in the photoelectric effect. It represents the minimum energy needed to release an electron from the surface of a material.

This intrinsic property is specific to each material and plays a vital role in determining whether light of a particular frequency can cause electron emission.
The work function (\( \Phi \)) is often measured in electron volts (eV) and can be converted to joules for calculations, especially when using other Standard International units.

• The relationship of work function in photoelectric experiments can be seen in the formula: \[ \Phi = h u - K \]
  Where:
  \( h \) is Planck’s constant and
  \( u \) is the frequency of incident light.
  \( K \) is typically the kinetic energy of the electrons once released.

The understanding of work function helps in assessing how different materials will react to incoming light and whether the energy they absorb will exceed this threshold, thus leading to the ejection of electrons, which is central to observations like those in this exercise.