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

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

Hydrogen and helium atoms are at temperatures of $27^{\circ}C$ and $127^{\circ}C$ respectively. Find the ratio of their de Broglie wavelengths.

The magnitude of the de-Broglie wavelength $(\lambda)$ of electron $(e)$,proton $(p)$,neutron $(n)$ and $\alpha-$ particle $(\alpha)$ all having the same energy of $1\,MeV$,in the increasing order will follow the sequence

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

If the kinetic energies of an electron, an alpha particle, and a proton having the same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$, and $\varepsilon_3$ respectively, then:

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 $nm$)