Problem 54 - Master Resource Book in JEE Main Physics

Question

What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength \(5200 \AA\) ? (a) \(10^{3} \mathrm{~ms}^{-1}\) (b) \(1.2 \times 10^{3} \mathrm{~ms}^{-1}\) (c) \(1.4 \times 10^{3} \mathrm{~ms}^{-1}\) (d) \(2.8 \times 10^{3} \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The electron's velocity should be \(1.4 \times 10^{3} \, \text{ms}^{-1}\). Option (c) is correct.

Step 1: Find Photon's Momentum

To find the momentum of the photon, use the formula \( p = \frac{h}{\lambda} \), where \( h = 6.63 \times 10^{-34} \, \text{Js} \) is Planck's constant and \( \lambda = 5200 \, \text{\AA} = 5200 \times 10^{-10} \, \text{m} \) is the wavelength. \[ p = \frac{6.63 \times 10^{-34}}{5200 \times 10^{-10}} \approx 1.275 \times 10^{-27} \, \text{kg m/s} \]

Step 2: Use Electron Momentum Formula

The momentum of the electron is given by the formula \( p = mv \), where \( m = 9.11 \times 10^{-31} \, \text{kg} \) is the mass of the electron and \( v \) is its velocity.

Step 3: Equate Momentums

Since the electron's momentum equals the photon's momentum, set \( mv = \frac{h}{\lambda} \). Substitute the known values: \[ 9.11 \times 10^{-31} \cdot v = 1.275 \times 10^{-27} \]

Step 4: Solve for Electron's Velocity

Solve for \( v \) by dividing both sides by the electron's mass: \[ v = \frac{1.275 \times 10^{-27}}{9.11 \times 10^{-31}} \approx 1.4 \times 10^{3} \, \text{ms}^{-1} \]

Step 5: Determine Correct Option

Compare the calculated velocity with the options provided. The correct option is (c) \(1.4 \times 10^{3} \, \text{ms}^{-1}\).

Key Concepts

Photon Momentum

The idea of photon momentum might sound strange at first. Usually, when we think of momentum, we associate it with objects having mass. However, a photon, which is a particle of light, is unique because it has momentum despite having no mass. The momentum of a photon is given by the formula \( p = \frac{h}{\lambda} \). In this equation, \( h \) is Planck's constant \((6.63 \times 10^{-34} \, \text{Js})\), and \( \lambda \) is the wavelength of the photon.What makes photon momentum important is its role in phenomena like the photoelectric effect, where light hits a material and ejects electrons. Understanding photon momentum helps us grasp how energy and momentum are transferred from light to electrons. In exercises like the one we examined, calculating photon momentum allows us to compare it to the momentum of other particles, like electrons.

Electron Velocity

In the realm of quantum physics, electrons carry momentum similar to how we view macroscopic objects. The momentum \( p \) of an electron is the product of its mass \( m \) and velocity \( v \), expressed as \( p = mv \). In our original exercise, we aimed to find the velocity of an electron such that its momentum equaled that of a photon. By setting the electron's momentum equal to the computed momentum of the photon, we find the electron's velocity. This comparison highlights how quantum objects, even those as different as photons and electrons, can have equivalent properties like momentum, further illustrating the wave-particle duality inherent in quantum mechanics. In this case, solving \( v \approx 1.4 \times 10^{3} \, \text{ms}^{-1} \) allows us to understand how fast an electron needs to move to match the photon's momentum.

Planck's Constant

Planck's constant, denoted as \( h \), is pivotal in the study of quantum mechanics. With a value of \( h = 6.63 \times 10^{-34} \, \text{Js} \), it's a tiny number but one with colossal significance. It bridges the gap between the pond of classical physics and the ocean of quantum theory.Planck's constant appears in many foundational equations, like the one for photon momentum \( p = \frac{h}{\lambda} \) and the energy of a photon \( E = hf \), where \( f \) is the oscillation frequency. It quantifies the discrete nature of energy exchanges and was crucial in explaining the photoelectric effect, which classical theories couldn't fully describe.Using Planck's constant in calculations helps pinpoint the link between energy and frequency, or in the case of momentum and wavelength for photons. As seen in the original calculation, it's an essential parameter for determining photon momentum in many physics problems.