1.

Calculate the (a) Momentum, and (b) De-broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Answer»

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SOLUTION :Here accelerating potential difference `V=56V`
(a) `THEREFORE` Momentum of electron `p=SQRT(2mK)=sqrt(2mVe)=sqrt(2xx9.11xx10^(-31)xx56xx1.6xx10^(-19))`
`=4.04xx10^(-24)KG" ms"^(-1)`
(b) `therefore`de-Broglie wavelength `lamda=(h)/(p)=(6.63xx10^(-34))/(4.04xx10^(-24))=1.64xx10^(-10)m=0.164nm`.


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