1.

What is the de-Broglie-wavelength of the electron accelerated through a potential difference of 100 V ?

Answer»

`12.27Å`
`1.227Å`
`0.1227Å`
`0.001227Å`

Solution :de-Broglie wavelength of an electron is given by :
`LAMBDA=(h)/(mv)=(h)/(sqrt(2meV))=(12.27)/(sqrt(V))Å`
(h=Plank.s constant)
where m=mass of electron
E= electronic CHARGE and
V= potential difference with which electron is accelerated.
`lambda=(12.27)/(sqrt(100))Å=(12.27)/(10)Å`
`=1.227Å`


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