Problem 5 - Master Resource Book in JEE Main Physics

Question

The energy that should be added to an electron to reduce its de-Broglie wavelength from \(10^{-10} \mathrm{~m}\) to \(0.5 \times 10^{-10} \mathrm{~m}\), will be (a) four times the initial energy (b) thrice the initial energy (c) equal to the initial energy (d) twice the initial energy

Step-by-Step Solution

Answer

The energy needed is four times the initial energy (Option a).

Step 1: Understanding the de-Broglie Wavelength Formula

The de-Broglie wavelength equation is given by \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\) and \( p \) is the momentum of the electron. Since momentum \( p \) is also given by \( p = mv \) (mass times velocity), the kinetic energy can be expressed as \( E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m} \).

Step 2: Expressing Wavelengths in Terms of Momentum

For the initial wavelength \( \lambda_1 = 10^{-10} \text{ m} \), the momentum \( p_1 = \frac{h}{\lambda_1} \). Similarly, for the new wavelength \( \lambda_2 = 0.5 \times 10^{-10} \text{ m} \), the momentum \( p_2 = \frac{h}{\lambda_2} \).

Step 3: Calculating the Energies for Both Wavelengths

Calculate the initial energy \( E_{k1} = \frac{p_1^2}{2m} \) and the new energy \( E_{k2} = \frac{p_2^2}{2m} \). These energies correspond to the initial and new momenta calculated from the de-Broglie wavelengths.

Step 4: Compare New Energy with Initial Energy

Determine the relationship between \( E_{k2} \) and \( E_{k1} \). From \( p_2 = 2p_1 \) (since \( \lambda_2 = 0.5\lambda_1 \)), we find \( E_{k2} = \frac{(2p_1)^2}{2m} = 4 \times \frac{p_1^2}{2m} = 4E_{k1} \). The energy increases by four times.

Key Concepts

Momentum and Energy Relationship

Understanding the connection between momentum and energy is central to solving problems involving de-Broglie wavelengths. The de-Broglie wavelength formula tells us how the momentum of a particle relates to its wavelength: \( \lambda = \frac{h}{p} \). Because the wavelength is inversely proportional to momentum, a decrease in wavelength results in an increase in momentum, assuming Planck's constant is unchanged. This is crucial when analyzing how changing the wavelength influences energy.
To comprehend the energy involved, remember that momentum \( p \) is defined as \( p = mv \) (mass times velocity). Kinetic energy, which is the energy of motion, can be expressed using momentum as \( E_k = \frac{p^2}{2m} \). Therefore, if we know how momentum changes, we can also deduce how kinetic energy changes. As in the exercise, if the momentum doubles (\( p_2 = 2p_1 \)), the kinetic energy changes accordingly, yielding an increase by the square factor: \( E_{k2} = 4E_{k1} \). This illustrates how significant the relationship between momentum and energy is in quantum mechanics.

Planck's constant

Planck's constant \( h \) is a fundamental quantity in the world of quantum physics, notable for its role in the de-Broglie wavelength formula. Its value remains steadfast at \( 6.626 \times 10^{-34} \text{ J s} \). This constant provides the bridge between the classical and quantum realms as it appears in equations linking wave-like and particle-like properties.
In the context of the de-Broglie wavelength, Planck's constant determines how small or large the wavelength will be for a given momentum. Because it is so minute, it magnifies the wave characteristics of small particles like electrons, making the wave-particle duality observable at a quantum level. Without this constant, the relation \( \lambda = \frac{h}{p} \) would lack the requisite balance between energy scales, rendering it essential for calculations involving energy and momentum.

Kinetic Energy Calculation

Kinetic energy is the energy possessed by an object due to its motion, and calculating it accurately is key to solving physics problems involving moving particles. For subatomic particles like electrons, kinetic energy can be fully understood through the relationship \( E_k = \frac{p^2}{2m} \), connecting energy to momentum.
When the de-Broglie wavelength shifts, the associated momentum and thus kinetic energy shifts as well. For example, halving the wavelength (\( \lambda_2 = 0.5 \lambda_1 \)) as in our exercise, doubles the momentum \( p_2 = 2p_1 \), and since kinetic energy is proportional to the square of the momentum, the kinetic energy quadruples: \( E_{k2} = 4E_{k1} \).

• This highlights the quadratic relationship between kinetic energy and momentum, resulting in significant energy changes with even small alterations in velocity or mass.
• These calculations are pivotal in quantum mechanics scenarios, where predicting particle behavior relies on understanding how these energies change with motion.

Understanding these concepts is vital for comprehending how particles behave under various energy inputs and calculating the necessary energy for achieving specific de-Broglie wavelengths.