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Calculate the de Broglie wavelength of a...

Calculate the de Broglie wavelength of an electron accelerating in a particle accelerator through a potential difference of 110 million volt.

Text Solution

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Kinetic energy of an electron under the potential difference of 110 million volt=110MeV=`110xx10^(6)eV`
`therefore (1)/(2)mv^(2)=110xx10^(6)eV=110xx10^(6)xx1.602xx10^(-19)J`
or, `(1)/(2)xx9.11xx10^(-31)xxv^(2)=110xx10^(6)xx1.602xx10^(-19)`
`therefore v=((2xx110xx10^(6)xx1.602xx10^(-19))/(9.11xx10^(-31)))^(1//2)`
`=6.22xx10^(9)m*s^(-1)`
`therefore lamda=(h)/(mv)=(6.626xx10^(-34))/((9.11xx10^(-31))xx(6.22xx10^(9)))`
`=1.17xx10^(-13)m`
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