Question

What is the De Broglie wavelength of an electron under 150 V acceleration?

Answer 1

0.1 nm

Step-by-step explanation:

Potential difference of the electron = 150 V

Mass of electron
`m=9.1* 10^(-31)kg`

Charge on electron
`1.6* 10^(-19)C`

Plank's constant
`h=6.67* 10^(-34)`

If the velocity of the electron is v

Then according to energy conservation
`eV =(1)/(2)mv^2`

`v=\sqrt{(2eV)/(m)}=\sqrt{(2* 1.6* 10^(-`19)* 150)/(9.1* 10^(-31))}=7.2627* 10^(6)m/sec`

According to De Broglie
`\lambda =(h)/(mv)=(6.67* 10^(-34))/(9.1* 10^(-31)* 7.2627* 10^(6))=0.1nm`