The energy that should be added to an electro | vedclass

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

Question

(D) We know that the de-Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.This implies th

Vedclass · Accepted Answer

three-times the initial energy

Vedclass · Answer

(D) We know that the de-Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.This implies that $K \propto \frac{1}{\lambda^2}$.Let $K_1$ be the initial kinetic energy at $\lambda_1 = 1 \,nm$ and $K_2$ be the final kinetic energy at $\lambda_2 = 0.5 \,nm$.Then, $\frac{K_2}{K_1} = \left( \frac{\lambda_1}{\lambda_2} \right)^2 = \left( \frac{1}{0.5} \right)^2 = 2^2 = 4$.So, $K_2 = 4K_1$.The energy that should be added to the electron is $\Delta K = K_2 - K_1 = 4K_1 - K_1 = 3K_1$.Therefore, the energy to be added is three-times the initial energy.

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \,nm$ to $0.5 \,nm$ is

Similar Questions

An electron microscope uses which property of the electron?

$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is:

$A$ proton and an electron are accelerated by the same potential difference. If their de-Broglie wavelengths are $\lambda_p$ and $\lambda_e$ respectively,then:

The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module