An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $λ$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2λ$. The value of $(V_1/V_2)$ is equal to

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: Plan:
Use the momentum picture.

Step 2: Steps:
$\lambda = \frac{h}{p}$, so doubling $\lambda$ halves $p$. Kinetic energy $= \frac{p^2}{2m}$ becomes one quarter, and since $eV$ is the kinetic energy, $V_2 = \frac{V_1}{4}$. So $\frac{V_1}{V_2} = 4$.

Final Answer:
The ratio $V_1:V_2$ is $4:1$, option (D). \[ \boxed{4:1} \]

An electron accelerated through a potential difference \(V_1\) has a de-Broglie wavelength of \(λ\). When the potential is changed to \(V_2\), its de-Broglie wavelength increases to \(2λ\). The value of \((V_1/V_2)\) is equal to

Show Hint

The Correct Option is D

Solution and Explanation

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