The de-Broglie wavelength of proton `(" charge" = 1.6 xx 10^(-19)C, "mass" = 1.6 xx 10^(-27) Kg)` accelerated through a potential difference of 1kV is

Options

A. `600 Å`
B. `0.9xx10^(-12)m`
C. `7 Å`
D. 0.9 nm

Answer

b

Explanation

de =Broglie wavelength ,
`lambda=(h)/(p)=(h)/(sqrt(2mE))=(h)/(sqrt(2mqV))`
`rArrlambda=(6.6xx10^(-34))/(sqrt(2xx1.6xx10^(-27)xx1.6xx10^(-19)xx1000))`
`rArrlambda=(6.6xx10^(-34))/(7.16xx10^(-22))=0.9xx10^(-12)m`