An electron has kinetic energy 2.8 times 10^{ | vedclass

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

Question

(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.Since kinetic energy $KE = \frac{p^2}{2m}$,we have $p = \sqrt{

Vedclass · Accepted Answer

$9.28 	imes 10^{-8} \ m$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.Since kinetic energy $KE = \frac{p^2}{2m}$,we have $p = \sqrt{2mKE}$.Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mKE}}$.Given $h = 6.62 	imes 10^{-34} \ J \cdot s$,$m = 9.1 	imes 10^{-31} \ kg$,and $KE = 2.8 	imes 10^{-23} \ J$.$\lambda = \frac{6.62 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 2.8 	imes 10^{-23}}}$.$\lambda = \frac{6.62 	imes 10^{-34}}{\sqrt{50.96 	imes 10^{-54}}} = \frac{6.62 	imes 10^{-34}}{7.138 	imes 10^{-27}} \approx 9.28 	imes 10^{-8} \ m$.

An electron has kinetic energy $2.8 \times 10^{-23} \ J$. The de-Broglie wavelength will be nearly $(m_e = 9.1 \times 10^{-31} \ kg)$.

Similar Questions

If the kinetic energy of an electron is $2.8 \times 10^{-23} \ J$,then the de Broglie wavelength will be ....... $(m_e = 9.1 \times 10^{-31} \ kg)$.

Calculate the de Broglie wavelength of a $CO_2$ molecule moving with a velocity of $440 \, m/s$. (Molar mass of $CO_2 = 44 \, g/mol$)

If the wavelength of a photon is $2.2 \times 10^{-11} \ m$ and $h = 6.6 \times 10^{-34} \ J \ s$,then the momentum of the photon is:

If the radius of the first orbit of $H$ atom is $a_0$, what is the de-Broglie wavelength of an electron in the third orbit (in $\pi a_0$)?

What is the mass of a particle with a wavelength of $3.313 \mathring{A}$ moving with a speed of $2.0 \times 10^8 \ m \ s^{-1}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module