An electron is moving with initial velocity $\overset{⃗}{v} = V_0\hat{j}$ in a magnetic field $\overset{⃗}{B} = B_0\hat{i}$. Its de-Broglie wavelength will

Question

A photon and an electron have the same energy $ E $. If $ \lambda_p $ is the wavelength of the photon and $ \lambda_e $ is the de Broglie wavelength of the electron, then the ratio $ \frac{\lambda_p}{\lambda_e} $ is:

Answer 1

Step 1: Approach
Use energy.

Step 2: Work-energy
Work done by a magnetic force is $\int\vec F\cdot d\vec s=0$ because $\vec F\perp\vec v$. So kinetic energy $\tfrac12mv^2$ does not change.

Step 3: Wavelength
$\lambda=\dfrac{h}{\sqrt{2mK}}$ with constant $K$, so $\lambda$ is constant. Option (C).

Final Answer:
A magnetic force does no work, so the speed and de Broglie wavelength stay constant, option (C). \[ \boxed{\text{Remains constant}} \]

An electron is moving with initial velocity \(\overset{⃗}{v} = V_0\hat{j}\) in a magnetic field \(\overset{⃗}{B} = B_0\hat{i}\). Its de-Broglie wavelength will

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The Correct Option is C

Solution and Explanation

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