The de-Broglie wavelength associated with an | vedclass

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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

$0.112$

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(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 $nm$)

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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 $nm$)