Question

Kinetic energy of an electron with de broglie wavelength 0.3nm is

Answer 1

we know that

$kinetic \: energy = \frac{ {h}^{2} }{2m { \lambda}^{2} } \\ where \: \: \: m = mass \: of \: electron \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = plancks \: constant \\ and \: \: \: \: \: \: \: \: \lambda = de \: broglie \: wavelength \\ therefore \\ \\ k.e = \frac{{(6.626 \times {10}^{ - 34}) }^{2} }{2 \times 9.1 \times {10}^{ - 31} \times {(0.3 \times {10}^{ - 9} )}^{2} } \\ \\ k.e = 2.68 \times {10}^{ - 18} joules = 16.72ev$