The kinetic energy of one electron is `2.8xx10^(-13)` J. What is the de-Broglie wavelength

Options

A. `9.25xx10^(-13)` m
B. `9.25xx10^(-16)` m
C. `9.25xx10^(-8)` m
D. `18.5xx10^(-13)` m

Answer

A

Explanation

`because KE=1/2mv^2`
Here, `m=9.1xx10^(-31)`kg, KE=`2.8xx10^(-13)`J
`therefore 2.8xx10^(-13)=1/2xx9.1xx10^(-31)xxv^2`
`v=sqrt((2xx2.8xx10^(-13))/(9.1xx10^(-31)))`
`v=0.784xx10^9` m/s
`because lambda=h/"mv"=(6.6xx10^(-34))/(9.1xx10^(-31)xx0.784xx10^9)`
`=0.925xx10^(-12)` m
`=9.25xx10^(-13) m `