The de-Broglie wavelength of an electron of kinetic energy 9 eV is (take, `h=4xx10^(-15)eV-s,c=3xx10^(10)cm//s` and the mass `m_(e)` of electron as `m_(e)c^(2)=0.5` MeV 

Options

A. `4xx10^(-8)` cm
B. `3xx10^(-8)` cm
C. `4xx10^(-7)` cm
D. `3xx10^(-7)` cm

Answer

A

Explanation

For electron, `pc=sqrt(2KEm_(0)c^(2))`
`implies" "pc=sqrt(2xx9eV xx0.58xx10^(6))`
`implies" "pc=3xx10^(3)eV`
Now, de-Broglie wavelength,
`lambda=(hc)/(pc)=(4xx10^(-15)xx3xx10^(10))/(3xx10^(3))=4xx10^(-8)" cm"`
`(because c=3xx10^(10)" cm"//"s")`