The de-Broglie wavelength associated with an | vedclass

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

Question

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqr

Vedclass · Accepted Answer

$0.112$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqrt{2meV}}$Given:$h = 6.6 	imes 10^{-34} \ J \cdot s$$m = 9 	imes 10^{-31} \ kg$$e = 1.6 	imes 10^{-19} \ C$$V = 121 \ V$Substituting the values:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 1.6 	imes 10^{-19} 	imes 121}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{3484.8 	imes 10^{-50}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{59.03 	imes 10^{-25}}$$\lambda \approx 0.1118 	imes 10^{-9} \ m = 0.112 \ nm$Thus,the correct option is $B$.

The de-Broglie wavelength associated with an electron,accelerated through a potential difference of $121 \ V$ is about:
[Take Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$,mass of electron $m = 9 \times 10^{-31} \ kg$,charge of electron $e = 1.6 \times 10^{-19} \ C$] (in $nm$)

Similar Questions

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is

According to the de-Broglie hypothesis, the ratio of the wavelength of a photon and that of an electron having the same energy $E$ is $(m = \text{mass of electron}, c = \text{velocity of light})$

If $\lambda$ and $K$ are the de Broglie wavelength and kinetic energy,respectively,of a particle with constant mass,the correct graphical representation for the particle will be:

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The de-Broglie wavelength associated with an electron,accelerated through a potential difference of $121 \ V$ is about:[Take Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$,mass of electron $m = 9 \times 10^{-31} \ kg$,charge of electron $e = 1.6 \times 10^{-19} \ C$] (in $nm$)