The potential difference V required for accel | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqr

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqrt{2meV}}$Squaring both sides, we get:$\lambda^2 = \frac{h^2}{2meV}$Rearranging for $V$:$V = \frac{h^2}{2me\lambda^2}$Using the standard approximation formula for an electron:$V \approx \frac{150}{\lambda^2} 	ext{ V}$, where $\lambda$ is in $	ext{Å}$.Given $\lambda = 1 	ext{ Å}$:$V = \frac{150}{(1)^2} = 150 	ext{ V}$.Thus, the required potential difference is $150 	ext{ V}$.

The potential difference $V$ required for accelerating an electron to have the de-Broglie wavelength of $1 \text{ Å}$ is (in $\text{ V}$)

Similar Questions

If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is

Two electrons are moving with the same speed $v$. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field. After some time,if the de-Broglie wavelengths of the two are $\lambda_1$ and $\lambda_2$,then:

$A$ photon and an electron are given the same energy $(10^{-20} \ J)$. If the wavelengths associated with the photon and the electron are $\lambda_{Ph}$ and $\lambda_{el}$ respectively,then which of the following statements is correct?

If the kinetic energy of an electron and a proton are equal,find the ratio of their associated de Broglie wavelengths.

Explain the particle-wave (dual) nature of matter.

Vedclass Test Series

Exam Paper Generator

Online Exam Module