Calculate the de Broglie wavelength of an ele | vedclass

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

Question

(A) The de Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$.Given:Planck's constant $(h)$ = $6.626 	imes 10^{-34} 	e

Vedclass · Accepted Answer

$3.31 	imes 10^{-10} 	ext{ m}$

Vedclass · Answer

(A) The de Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$.Given:Planck's constant $(h)$ = $6.626 	imes 10^{-34} 	ext{ J s}$.Mass of electron $(m)$ = $9.1 	imes 10^{-31} 	ext{ kg}$.Velocity of electron $(v)$ = $2.2 	imes 10^6 	ext{ m s}^{-1}$.Substituting these values into the formula:$\lambda = \frac{6.626 	imes 10^{-34}}{(9.1 	imes 10^{-31}) 	imes (2.2 	imes 10^6)}$.$\lambda = \frac{6.626 	imes 10^{-34}}{20.02 	imes 10^{-25}}$.$\lambda = 0.33096 	imes 10^{-9} 	ext{ m}$.$\lambda = 3.31 	imes 10^{-10} 	ext{ m}$.Thus, the correct option is $A$.

Calculate the de Broglie wavelength of an electron in the first Bohr orbit of a hydrogen atom if the velocity of an electron in the first orbit is $2.2 \times 10^6 \text{ m s}^{-1}$. [mass of electron = $9.1 \times 10^{-31} \text{ kg}$, Planck's constant $(h)$ = $6.626 \times 10^{-34} \text{ J s}$]

Similar Questions

If $|\vec{A} + \vec{B}| = |\vec{A}| + |\vec{B}|$,the angle between $\vec{A}$ and $\vec{B}$ is ....... $^o$.

$\vec{A}$ and $\vec{B}$ are two non-zero vectors inclined at an angle $\theta$. $\hat{a}$ and $\hat{b}$ are unit vectors along $\vec{A}$ and $\vec{B}$ respectively. The component of $\vec{A}$ in the direction of $\vec{B}$ is

The angles which a vector $\hat{i} + \hat{j} + \sqrt{2} \hat{k}$ makes with $X, Y$ and $Z$ axes respectively are

Given that $\overrightarrow A + \overrightarrow B + \overrightarrow C = 0$. Out of three vectors,two are equal in magnitude and the magnitude of the third vector is $\sqrt{2}$ times that of either of the two having equal magnitude. Then the angles between the vectors are given by:

If $\alpha, \beta$ and $\gamma$ are the angles made by a vector with $x, y$ and $z$ axes respectively,then $\sin ^2 \alpha + \sin ^2 \beta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module