The de-Broglie wavelength of a body of mass 1 | 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: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 1 \ kg$,$v

Vedclass · Accepted Answer

$3.3 	imes 10^{-27} \ \mathring{A}$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 1 \ kg$,$v = 2000 \ m/s$.Substituting the values: $\lambda = \frac{6.63 	imes 10^{-34}}{1 	imes 2000} = \frac{6.63}{2} 	imes 10^{-37} \ m = 3.315 	imes 10^{-37} \ m$.Since $1 \ \mathring{A} = 10^{-10} \ m$,we convert the wavelength to $\mathring{A}$:$\lambda = 3.315 	imes 10^{-37} \ m = 3.315 	imes 10^{-27} \ \mathring{A}$.Rounding to one decimal place,we get $3.3 	imes 10^{-27} \ \mathring{A}$.

The de-Broglie wavelength of a body of mass $1 \ kg$ moving with a velocity of $2000 \ m/s$ is:

Similar Questions

What is the $(a)$ momentum,$(b)$ speed,and $(c)$ de Broglie wavelength of an electron with kinetic energy of $120 \ eV$?

When the kinetic energy of an electron is increased,the wavelength of the associated wave will:

Electrons are accelerated through a potential difference of $16 \ kV$. If the potential difference is increased to $64 \ kV$,then the de-Broglie wavelength associated with the electron will

An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75 \%$ of the initial value?

Vedclass Test Series

Exam Paper Generator

Online Exam Module