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De - Broglie wavelength of an electron a...

De - Broglie wavelength of an electron accelerated by a voltage of 50 V is close to
`(|e|=1.6xx10^(-19)C,m_(e)=9.1xx10^(-31))`

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The correct Answer is:
C
`lamda=(12.27)/sqrtVÅ=(12.27)/sqrt50Å=1.73Å`
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de-Brogile wavelength of an electron accelerated by a voltage of 50 V is close to ( |e| = 1.6 xx 10^(-19)C, m_(e) = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) Js) :-

Calculate the (a) momentum and (b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56V. Given, h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg, e=1.6xx10^(-19)C .

The de Broglie wavelength of an electron with kinetic energy 120 e V is ("Given h"=6.63xx10^(-34)Js,m_(e)=9xx10^(-31)kg,1e V=1.6xx10^(-19)J)

What is the (a) momentum (b) speed and (c) de-Broglie wavelength of an electron with kinetic energy of 120 eV. Given h=6.6xx10^(-34)Js, m_(e)=9xx10^(-31)kg , 1eV=1.6xx10^(-19)J .

What is the approximate value of the de broglie wavelength of an electron having 80 eV of electron ? (1eV = 1.6xx10^(-19) J " mass of electron " = 9xx10^(-31) kg, " Plank's constant " = 6.6xx10^(-34) J-sec)

The de Broglie wavelength of an electron in a metal at 27^(@)C is ("Given"m_(e)=9.1xx10^(-31)kg,k_(B)=1.38xx10^(-23)JK^(-1))

if the de broglie wavelength of an electron is 0.3 nanometre, what is its kinetic energy ? [h=6.6xx10^(-34) Js, m=9xx10^(-31)kg, 1eV = 1.6xx10^(-19) J]

The de - Broglie wavelength of an electron having 80 ev of energy is nearly ( 1eV = 1.6 xx 10^(-19) J , Mass of electron = 9 xx 10^(-31) kg Plank's constant = 6.6 xx 10^(-34) J - sec )

The de-Broglie wavelength of an electron is 600 nm . The velocity of the electron is: (h = 6.6 xx 10^(-34) J "sec", m = 9.0 xx 10^(-31) kg)

Calculate the de-Broglie wavelength of an electron of kinetic energy 100 eV. Given m_(e)=9.1xx10^(-31)kg, h=6.62xx10^(-34)Js .

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