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

(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.This implies $\la

Vedclass · Accepted Answer

Thrice the initial energy

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.This implies $\lambda \propto \frac{1}{\sqrt{E}}$,or $E \propto \frac{1}{\lambda^2}$.Let the initial wavelength be $\lambda_1 = 10^{-10} \ m$ and the final wavelength be $\lambda_2 = 0.5 	imes 10^{-10} \ m$.Then,$\frac{E_2}{E_1} = \left( \frac{\lambda_1}{\lambda_2} ight)^2 = \left( \frac{10^{-10}}{0.5 	imes 10^{-10}} ight)^2 = (2)^2 = 4$.Thus,the final energy $E_2 = 4E_1$.The energy that should be added is $\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$.

The energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be

Similar Questions

Let $f(x)=15-|x-10| ; x \in R$. Then,the set of all values of $x$,at which the function $g(x)=f(f(x))$ is not differentiable,is

Boric acid is $NOT$ used in which of the following?

The total number of possible isomers of the compound $[Cu(NH_3)_4][PtCl_4]$ are:

Which of the following is the correct sequence of the various steps of Herbarium formation?
$1$ Drying $2$ Poisoning $3$ Collection $4$ Labelling $5$ Mounting $6$ Deposition $7$ Stitching

The angle between the lines joining the origin to the points of intersection of the line $x + 2y + 1 = 0$ and the curve $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module