1.

What is the de Broglie length of an electron,which is accelerated from the rest,through a potential difference of 100v?

Answer»

Solution :Potential difference =b=100v
Potential energy=eV=`1.609xx10^(-19)cxx100`
`1/2mv^(2)=1.609xx10^(-19)XX100`
`1/2mv^(2)=1.609xx10^(17)J`
`v^(2)=(2xx1.609xx10^(-10))/(m)`
m=mass of elctron`=9.1xx10^(-31)KG`
`:.v^(2)=(2xx1.609xx10^(17))/(9.1xx10^(-31))`
`:.v=SQRT(2xx1.609xx10^(-17))/(9.1xx10^(31))`=`sqrt(2xx1.609xx10^(-17)xx10^(31)/(9.1)`
`v=sqrt(3.218xx10^(14)/(9.1)`
`v=5.93xx10^(6)m/s`
`gamma=h/(MV) Where h=6.62xx10^(-34)JS`
`=(6.62xx10^(-34))/(9.1xx10^(-31)xx5.93xx10^(6))`
`1.2xx10^(-10)m`
`gamma=1.2A`


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