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 $A$.

Column $-I$	Column $-II$
$(1)$ Angular momentum	$(a)$ Scalar
$(2)$ Potential energy	$(b)$ Vector
	$(c)$ Unit vector

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

Which of the following quantity/quantities are dependent on the choice of orientation of the co-ordinate axes?
$(a)$ $\vec{a}+\vec{b}$
$(b)$ $3a_x+2b_y$
$(c)$ $(\vec{a}+\vec{b}-\vec{c})$

Match Column $-I$ with Column $-II$.

Column $-I$ Column $-II$

$(1)$ Angular momentum $(a)$ Scalar

$(2)$ Potential energy $(b)$ Vector

$(c)$ Unit vector

The vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ lie in a plane. Another vector $\overrightarrow{C}$ lies outside this plane. The resultant $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}$ of these three vectors:

The unit vector along $\hat{i} + \hat{j}$ is

$A$ vector has a magnitude the same as that of $\overrightarrow{A} = 3 \hat{i} + 4 \hat{j}$ and is parallel to $\overrightarrow{B} = 4 \hat{i} + 3 \hat{j}$. The $x$ and $y$ components of this vector in the first quadrant are $X$ and $3$ respectively,where $X = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module