Question

An electron accelerated by voltage, v = 100 volt, inside an electron gun, then what is the de-broglie wavelength of electron when it emerges from gun

Answer 1

lamda=√150/100=√1.5......

Answer 2

Answer:

An electron accelerated by voltage, v = 100 v inside an electron gun, then the de-Broglie wavelength of an electron when it emerges from the gun is $0.22*10^{-10} m$

Explanation:

The de Broglie wavelength is the wavelength of particle-wave (dual nature of matter that is when a particle is moving with a velocity sometimes it acts as a particle or wave)

The wavelength of a particle-wave is λ $=h/p$

where Planck's constant $h=6.26*10^{-34}Js$ and $p$- momentum of the particle

when a charged particle is accelerated by a voltage $V$

the de Broglie wavelength is λ= $h/\sqrt{2mqV}$

here the given particle is an electron

⇒ mass of an electron $m=9.1*10^{-31}kg$

charge of the electron $q=1.6*10^{-19}C$

accelerated voltage $V=100v$

substitute the given values to get the wavelength of electron wave

λ=$\frac{6.626*10^{-34} }{\sqrt{2*9.1*10^{-31}*1.6*10^{-19}*100 } }$

 =$0.22*10^{-10} m$