Calculate the (1) de Broglie wavelength of electron (2) momentum & (3) speed which has kinetic energy = 120eV

Explanation

We know that
(i) Linear momentum
`p = mv = h//lambda`
`= (6.625xx10^(-34))/(1.120xx10^(-10))`
`p = 5.915 xx10^(-24)kgms^(-1)`
(ii) Speed of electron `=v = sqrt((2KE)/m)`
i.e., `v=sqrt((2xx120xx1.6xx10^(-19))/(9.1xx10^(-31)))`
`v = 6.496 xx 10^(6)ms^(-1)`
(iii) `lambda = (12.27)/sqrt(v) `Å
given `KE = 120 =eV`
`therefore v = 120 V`
i.e., `lambda = (12.27)/sqrt(120)` Å
`lambda = 1.20 xx 10^(-10) m = 1.120` Å