Problem 52
Question
An electron (mass \(m\) ) with an initial velocity \(\mathbf{v}=v_{0} \hat{\mathbf{i}}\) is in an electric \(\mathbf{E}=E_{0} \hat{\mathbf{j}} .\) If \(\lambda_{0}=h / m v_{0}\), it's de-Broglie wavelength at time \(t\) is given by \(\quad\) (a) \(\lambda_{0}\) (b) \(\lambda_{0} \sqrt{1+\frac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}\) (c) \(\frac{\lambda_{\mathrm{D}}}{\sqrt{1+\frac{\mathrm{e}^{2} E_{0^{2}}^{2}}{m^{2} v_{0}^{2}}}}\) (d) \(\frac{\lambda_{0}}{\left(1+\frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}\right)}\)
Step-by-Step Solution
Verified Answer
Option (c) is correct: \( \frac{\lambda_0}{\sqrt{1 + \frac{e^2E_0^2t^2}{m^2v_0^2}}} \).
1Step 1: Understand the Scenario
An electron is moving with an initial velocity \( \mathbf{v}=v_0 \hat{\mathbf{i}} \) in an electric field \( \mathbf{E}=E_0 \hat{\mathbf{j}} \). We need to calculate its de-Broglie wavelength, \( \lambda(t) \), at time \( t \).
2Step 2: Analyze Force and Motion
The electric field \( \mathbf{E} = E_0 \hat{\mathbf{j}} \) exerts a force \( \mathbf{F} = eE_0 \hat{\mathbf{j}} \) on the electron. Since force equals mass times acceleration \( \mathbf{F} = m \mathbf{a} \), the acceleration \( \mathbf{a} = \frac{eE_0}{m} \hat{\mathbf{j}} \). This affects the velocity in the \( \hat{j} \) direction but not in the \( \hat{i} \) direction.
3Step 3: Compute Velocity over Time
The initial velocity is \( v_0 \hat{\mathbf{i}} \). Over time, due to the force, the velocity in the \( \hat{\mathbf{j}} \) direction increases as \( v_j = \frac{eE_0}{m}t \). Thus, the velocity at any time \( t \) is \( \mathbf{v}(t) = v_0 \hat{\mathbf{i}} + \frac{eE_0}{m}t \hat{\mathbf{j}} \).
4Step 4: Calculate Speed at Time t
The speed \( v(t) \) is given by the magnitude of \( \mathbf{v}(t) \). Compute it as: \[ v(t) = \sqrt{v_0^2 + \left( \frac{eE_0}{m}t \right)^2}\] This gives the resultant speed of the electron at time \( t \).
5Step 5: Determine de-Broglie Wavelength
The de-Broglie wavelength \( \lambda(t) \) is given by \( \lambda(t) = \frac{h}{mv(t)} \). Substituting the expression for speed, we find:\[ \lambda(t) = \frac{h}{m \sqrt{v_0^2 + \left( \frac{eE_0}{m}t \right)^2}}\]Factor \( v_0^2 \) out of the square root to simplify:\[ \lambda(t) = \frac{h}{m v_0} \frac{1}{\sqrt{1 + \frac{e^2E_0^2t^2}{m^2v_0^2}}}\]Substitute \( \lambda_0 = \frac{h}{mv_0} \):\[ \lambda(t) = \frac{\lambda_0}{\sqrt{1 + \frac{e^2E_0^2t^2}{m^2v_0^2}}}\]
6Step 6: Match with Given Options
The expression matches option (c): \( \frac{\lambda_0}{\sqrt{1 + \frac{e^2E_0^2t^2}{m^2v_0^2}}} \). This confirms the correct de-Broglie wavelength formula at time \( t \).
Key Concepts
Electron Motion in Electric FieldVelocity CalculationAcceleration Due to Electric FieldForce and Motion in Physics
Electron Motion in Electric Field
When an electron moves through an electric field, its motion is influenced by the electric forces acting upon it. In this scenario, the electron initially travels with velocity \( \mathbf{v} = v_0 \hat{\mathbf{i}} \), parallel to the x-axis. Meanwhile, the electric field \( \mathbf{E} = E_0 \hat{\mathbf{j}} \) is directed along the y-axis, perpendicular to the initial motion of the electron.
An electric field consists of lines of force exerting a push or pull on charged particles, like electrons. In our case, the field applies a force in the y-direction. The electron experiences this as a change in velocity along the y-axis, resulting in a change of motion over time.
An electric field consists of lines of force exerting a push or pull on charged particles, like electrons. In our case, the field applies a force in the y-direction. The electron experiences this as a change in velocity along the y-axis, resulting in a change of motion over time.
- The initial motion does not change along the x-axis.
- The electric field adds a component of velocity along the y-axis.
Velocity Calculation
Velocity is the rate of change of an object's position with respect to time. In this exercise, we start with the electron's initial velocity of \( v_0 \hat{\mathbf{i}} \), directed along the x-axis. As the electric field exerts a force\( \mathbf{F} = eE_0 \hat{\mathbf{j}} \), it leads to an additional velocity component along the y-axis.
The electron's velocity \( \mathbf{v}(t) \) at any time \( t \) becomes a vector with components in both the x and y directions:
The electron's velocity \( \mathbf{v}(t) \) at any time \( t \) becomes a vector with components in both the x and y directions:
- \( v_x(t) = v_0 \), remaining constant in the x-direction.
- \( v_y(t) = \frac{eE_0}{m}t \), developing in the y-direction over time.
Acceleration Due to Electric Field
Acceleration is a crucial concept in understanding how forces affect motion. It represents the rate of change of velocity over time. In the context of our exercise, the electric field influences the electron, providing an acceleration along the y-axis.
Given that the force exerted by the electric field is \( \mathbf{F} = eE_0 \hat{\mathbf{j}} \), we use Newton's second law of motion which states \( \mathbf{F} = m \mathbf{a} \). Here, \( e \) is the charge of the electron, \( E_0 \) is the magnitude of the electric field, and \( m \) is the mass of the electron.
Given that the force exerted by the electric field is \( \mathbf{F} = eE_0 \hat{\mathbf{j}} \), we use Newton's second law of motion which states \( \mathbf{F} = m \mathbf{a} \). Here, \( e \) is the charge of the electron, \( E_0 \) is the magnitude of the electric field, and \( m \) is the mass of the electron.
- The acceleration \( \mathbf{a} = \frac{eE_0}{m} \hat{\mathbf{j}} \) occurs only in the y-direction.
- There is no acceleration in the x-direction since the electric field is perpendicular to it.
Force and Motion in Physics
The relationship between force and motion is a fundamental element of physics, especially when analysing particle dynamics in electric fields. Newton’s laws of motion describe how forces influence the movement of objects. In this problem, the focus centers on how the electric force impacts an electron's motion, altering its velocity and, consequently, its de-Broglie wavelength over time.
The de-Broglie wavelength \( \lambda \) is intimately linked to a particle’s momentum, given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck’s constant and \( p \) the momentum. As velocity changes, so does momentum; hence the wavelength is influenced.
The de-Broglie wavelength \( \lambda \) is intimately linked to a particle’s momentum, given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck’s constant and \( p \) the momentum. As velocity changes, so does momentum; hence the wavelength is influenced.
- The electron begins with a momentum component \( mv_0 \).
- Over time, additional momentum develops due to changes from \( v_y(t) \).
Other exercises in this chapter
Problem 50
An electron (mass \(m\) ) with an initial velocity \(\mathbf{v}=v_{0} \hat{\mathbf{i}}\left(\mathbf{v}_{0}>0\right)\) is in an electric field \(\mathbf{E}=E_{0}
View solution Problem 51
If the mass of neutral \(=1.7 \times 10^{-27} \mathrm{~kg}\), then the de- Broglie wavelength of neutral of energy \(3 \mathrm{eV}\) is \(\left(h=6.6 \times 10^
View solution Problem 53
An electron of mass \(m\) and charge \(e\) initially at rest gets accelerated by a constant electric field \(E\). The rate of change of de-Broglie wavelength of
View solution Problem 54
What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength \(5200 \AA\) ? (a) \(10^{3} \mathrm{~ms}^{-1}\)
View solution