Problem 21 - Master Resource Book in JEE Main Physics

Question

For a certain metal, \(v=2 v_{0}\) and the electrons come out with a maximum velocity of \(4 \times 10^{6} \mathrm{~ms}^{-1}\). If the value of \(v=5 v_{0}\), then maximum velocity of photelectrons will be (a) \(2 \times 10^{7} \mathrm{~ms}^{-1}\) (b) \(8 \times 10^{6} \mathrm{~ms}^{-1}\) (c) \(2 \times 10^{6} \mathrm{~ms}^{-1}\) (d) \(8 \times 10^{5} \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The maximum velocity of the photoelectrons is \(8 \times 10^6\) ms\(^{-1}\), option (b).

Step 1: Understand the Photoelectric Effect Equation

The photoelectric effect equation relates the kinetic energy of ejected electrons to the energy of the incident light and the work function of the metal. It is given by \( KE_{ ext{max}} = h(v - v_0) \), where \( KE_{ ext{max}} = \frac{1}{2}mv^2 \), \( h \) is Planck's constant, \( v \) is the frequency of the incident light, \( v_0 \) is the threshold frequency, and \( v \) is the velocity of the electrons.

Step 2: Convert Given Information into Equation Form

Given \( v = 2v_0 \), and the maximum velocity is \( 4 \times 10^6 \) ms\(^{-1}\). We use the photoelectric equation: \( \frac{1}{2}mv^2 = h(2v_0 - v_0) \). Similarly, if \( v = 5v_0 \), we need to find the new maximum velocity, which will be \( \frac{1}{2}mv'^2 = h(5v_0 - v_0) \).

Step 3: Calculate Initial Kinetic Energy and Maximum Energy

For \( v = 2v_0 \), \( \frac{1}{2}m(4 \times 10^6)^2 = hv_0 \), providing the base kinetic energy. For \( v = 5v_0 \), \( \frac{1}{2}mv'^2 = 4hv_0 \), indicating the energy is now four times that initial value.

Step 4: Express New Velocity in Terms of Known Values

Given \( \frac{1}{2}mv'^2 = 4 \times \frac{1}{2}m(4 \times 10^6)^2 \), solve for \( v' \): \( v'^2 = 4 \times (16 \times 10^{12}) \). Simplifying, \( v' = \sqrt{64 \times 10^{12}} = 8 \times 10^6 \) ms\(^{-1}\).

Step 5: Identify the Correct Option

The maximum velocity of the photo-electrons when \( v = 5v_0 \) is \( 8 \times 10^6 \) ms\(^{-1}\), which matches option (b).

Key Concepts

Kinetic Energy

In the context of the photoelectric effect, kinetic energy plays a crucial role in understanding how electrons behave when they are ejected from a material. When light strikes a metal surface, it transfers energy to the electrons. If this energy is enough to overcome the work function of the metal, electrons are ejected with some kinetic energy.

The kinetic energy of the ejected electrons in a photoelectric experiment is calculated using the equation: \[KE_{\text{max}} = \frac{1}{2}mv^2,\] where:

• \( KE_{\text{max}} \) is the maximum kinetic energy of the electron,
• \( m \) is the mass of the electron, and
• \( v \) is the velocity of the ejected electron.

In this problem, we're given that the initial maximum velocity of ejected electrons is \( 4 \times 10^6 \) ms\(^{-1}\). From this, we compute the kinetic energy using the given formula. When conditions change, like varying light frequency, this kinetic energy changes accordingly. Understanding kinetic energy helps us relate the velocity of electrons to the frequency of incident light.

Planck's Constant

Planck's constant \( h \) is a fundamental constant in physics that relates the energy of a photon to the frequency of the light. In the photoelectric effect, it helps us determine how much energy is being transferred from the light to the electron.

The photoelectric equation, which holds for the scenario where light of frequency \( v \) hits a metal, is given by: \[KE_{\text{max}} = h(v - v_0),\] where:

• \( h \) is Planck's constant,
• \( v \) is the frequency of the incident light, and
• \( v_0 \) is the threshold frequency of the metal.

Planck's constant is pivotal because it quantifies the amount of energy a single photon carries at a specific frequency. By determining this, we can understand the energetics of the electron as it gains enough energy to escape from the material. Essentially, without Planck's constant, linking light properties with electron behavior would be impossible. In this exercise, \( h \) allows us to balance and solve our equation for electron velocity changes effectively.

Threshold Frequency

Threshold frequency \( v_0 \) is a vital concept when discussing the photoelectric effect as it represents the minimum frequency needed for electrons to be ejected from a metal surface.

The threshold frequency is the cut-off below which no electrons will be emitted, regardless of the light's intensity. It is a property unique to each material and influences whether electron ejection occurs. This can be seen in the equation for the photoelectric effect: \[KE_{\text{max}} = h(v - v_0).\]

• When \( v < v_0 \), \( KE_{\text{max}} \) becomes zero and no electron is ejected.
• When \( v \geq v_0 \), electrons are emitted and their kinetic energy can be calculated.

In this exercise, the threshold frequency serves as a reference point for calculating electron velocity. A higher frequency \( v \) provides the electrons with ample energy, ensuring they overcome the work function and are ejected at measurable velocities. Understanding threshold frequency helps explain why certain frequencies work to eject electrons while others do not.