The wavelength of a charged particle of mass `8.0xx10^(-31)`kg, charge `1.6xx10^(-19)C` and kinetic energy 3 keV will be ( Planck constant, `h=6.4xx10^(-34)Js` )

Options

A. `0.4 Å`
B. 2.1 Å
C. 1.0 Å
D. 1 Å

Answer

A

Explanation

Here, `m=8xx10^(-31)kg, q=1.6xx10^(-19)C`
and `KE=3` keV
Now, the de-Broglie wavelength, of a charged particle,
`lambda=h/p` ...(i)
and `p=sqrt(2m KE)` ...(ii)
From Eq. (i) and (ii) , we get
`lambda= (h)/(sqrt(2m KE))`
Putting the given values, we get
`rArr lambda=(6.4xx10^(-34))/(sqrt(2xx8xx10^(-31)xx3xx10^(3)xx1.6xx10^(-19)))`
`rArr lambda~~=0.4 Å`
Hence, the correct option is (a).