Problem 12 - Master Resource Book in JEE Main Physics

Question

An electron of mass \(m\) when accelerated through a potential difference has de-Broglie wavelength \(\lambda\). The de-Broglie wavelength associated with a proton of mass \(M\) accelerated through the same potential difference will be (b) \(\lambda \sqrt{\frac{m}{M}}\) (a) \(\lambda \frac{m}{M}\) (c) \(\lambda \frac{M}{m}\) (d) \(\lambda \sqrt{\frac{M}{m}}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(\lambda \sqrt{\frac{m}{M}}\).

Step 1: Write down the de-Broglie wavelength formula

The de-Broglie wavelength \( \lambda \) of a particle is given by: \[\lambda = \frac{h}{p}\]where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

Step 2: Relate momentum to kinetic energy

The kinetic energy \( KE \) given by the potential difference \( V \) is related to momentum by:\[ KE = \frac{1}{2}mv^2 = qV \] Thus, replacing \( mv \) using \( p = mv \), \[ p = \sqrt{2mqV} \]

Step 3: Determine de-Broglie wavelength for the electron

Substitute the electron's momentum expression into the wavelength formula:\[ \lambda_e = \frac{h}{\sqrt{2meqV}} \] This gives the wavelength for the electron.

Step 4: Determine de-Broglie wavelength for the proton

Using the same potential difference for the proton, which gives:\[ p = \sqrt{2MqV} \] The wavelength for the proton becomes:\[ \lambda_p = \frac{h}{\sqrt{2MqV}} \]

Step 5: Compare electron and proton wavelengths

Consider the fraction \( \frac{\lambda_p}{\lambda_e} \);\[ \frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2MqV}}}{\frac{h}{\sqrt{2meqV}}} \]This simplifies to:\[ \frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m}{M}} \]Thus, \( \lambda_p = \lambda_e \cdot \sqrt{\frac{m}{M}} \).

Step 6: Select the correct answer option

The relationship \( \lambda_p = \lambda \cdot \sqrt{\frac{m}{M}} \) corresponds to option (b) in the provided choices.

Key Concepts

Electron Acceleration

Understanding electron acceleration helps us grasp how these particles gain energy. When an electron is subjected to a potential difference, it experiences a force that causes it to accelerate. Electrons, being negatively charged, move toward the positive potential.
This acceleration results in an increase in kinetic energy, turning electrical potential energy into motion. The potential difference essentially acts like a speeding lane, giving the electron the energy it needs to move faster.
Imagine it as a downhill slope for a skateboarder; the steeper the slope, the faster they go. Similarly, the larger the potential difference, the greater the acceleration experienced by the electron.

Potential Difference

Potential difference, often referred to as voltage, is a measure of energy difference per unit charge between two points. In this context, it provides the force that accelerates the electron. For an electron, which starts from rest, this energy becomes kinetic as it moves under the influence of this potential.
Potential difference is crucial because it determines the amount of kinetic energy gained by the electron. Thus, an increase in potential difference would lead to a higher kinetic energy and consequently, a smaller de-Broglie wavelength.

• Voltage pushes the electron through an electric field.
• Kinetic energy increases with higher potential difference.
• Relates directly to the particle's momentum and therefore its wavelength.