An electron is accelerated by a potential difference of 50 V. Find the de Broglie wavelength associated with it.

Options

A. `2.7xx10^(-10)` m
B. `1.74xx10^(-10)` m
C. `3.6xx10^(-9)` m
D. `4.9xx10^(-11)` m

Answer

B

Explanation

The gain of kinetic energy by an electron is eV.
`(1)/(2)mv^2=eV`
`upsilon=sqrt((2eV)/(m))=sqrt((2(1.60xx10^(-19))(50))/((9.11xx10^(-31))))`
`=4.19xx10^(6)ms^(-1)`
Thus, the electron's de Broglie wavelength is `lamda=(h)/(mv)=(6.63xx10^(-34))/((9.11xx10^(-31))(4.19xx19^(6)))=1.74xx10^(-10)m`