The wavelength associated with an electron accelerated through a potential difference of `100 V` is nearly

Answer

`1.22xx10^(-13)m`

Explanation

Step I. Calculation of velocity of electron
Kinetic Energy (KE) of electron =100 million volts `=100xx10^(6)=10^(8)` volts
`=10^(8)xx1.6xx10^(-19)J=1.6xx10^(11) "kg" m^(2)s^(-2)`
`(1)/(2) mv^(2)=1.6xx10^(11) "kg" m^(2)s^(-2)`
`v=((2xx1.6xx10^(-11) "kg" m^(2)s^(-2))/((9.1xx10^(-31) "kg")))^(1//2)=5.93xx10^(9) m s^(-1)`
Step II. Calculation of wavelength of electron
According to de Broglie's equation,
`lambda=(h)/(mv)=((6.626xx10^(-34) "kg"m^(2)s^(-1)))/((9.1xx10^(-31)"kg")xx(5.93xx10^(9) ms^(-1)))=1.22xx10^(-13) m`