The de broglie wavelength of electron moving with kinetic energy of 144 eV is nearly

Answer

A, B, C, D

Explanation

`E = (1)/(2) mv^(2) = 100 eV = 100 xx 1.6 xx 10^(-19) J`
`v^(2)= (2E)/(m)`
`v = ((2E)/(m))^(1//2)`
`lambda = ((2E)/(m))^(1//2) = sqrt(2mE)`
`lambda = (h)/(mv) = (h)/(sqrt(2mE)0) meter `
`lambda = (6.6 xx 10^(-34))/((2 xx 9.1 xx 10^(-31) xx 100 xx 1.6 xx 10^(-19))^(1//2))`
`= 1.23 xx 10^(-10) m = 1.23 Å`