Through what potential difference must an electron be accelerated to have a wavelength of 632.8 nm corresponding to the wavelength of the red light of a He-Ne laser? A)3.77 * 10^-6 V B)5.56 * 10^-4 V C)1.25 * 10^-2 V D)3.77 V

Question

Quizplus · Accepted Answer

To find the potential difference required to accelerate an electron to a specific wavelength, we can use the de Broglie wavelength formula $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the electron. The kinetic energy acquired by the electron when accelerated through a potential difference $V$ is given by $eV = \frac{1}{2}mv^2$, which can be related to the momentum $p = mv$. By substituting and rearranging, we find that the potential difference $V$ is directly related to the desired wavelength. The correct calculation leads to the answer in option A, which is $3.77 \times 10^{-6} V$, but due to a formatting error in the options, it appears there might be a misunderstanding in the representation of the units or the magnitude. The correct magnitude and reasoning lead to the selection of option A as the closest match despite the apparent typographical errors in the options.

Through What Potential Difference Must an Electron Be Accelerated to Have