What amount of energy (in eV) should be added to an electron to reduce its de-Broglie its de-Broglie wavelength from 100 to 50 pm? Given `h=6.62xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg`.

Answer

`452eV`

Explanation

Here, `lambda_(1)=100 "pm"=100xx10^(-12)m`,
`lambda_(2)=50xx10^(-12)m`
De-Broglie wavelength,
`lambda=h/(mv) and K=1/2mv^(2)`
`:. lambda=h/(sqrt(2mK))`
In first case, `100xx10^(-12)=h/(sqrt(2mK_(1)))`........(i)
In second case, `50xx10^(-12)=h/(sqrt(2mK_(2)))`......(ii)
From (i) and (ii)
`2=sqrt((K_(2))/(K_(1))) or K_(2)=4K_(1)`
Energy to be added =`4K_(2)-K_(1)=3K_(1)`
From (i)
`sqrt(2mK_(1))=(6.63xx10^(-34))/(10^(-10))=6.63xx10^(-24)`
`K_(1)=((6.63xx10^(-34))^(2))/(2xx9.1xx10^(-31))J`
Energy added`=3K_(1)`
`=(3xx(6.63xx10^(-34))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19)) eV`
`=4.52xx10^(2)=452 eV`