Question

If the de Broglie wavelength of an electron is 258 nm , what is the kinetic energy of this electron?

Answer 1

Since, de Broglie wavelength() = h/(m*v)

also, kinetic energy(K) = 0.5*m*v^2

then,

= h/sqrt(2*m*K)

K = h^2/(2*m*^2)

here, h = 6.626*10-34

m = mass of electron = 9.1*10^-31 kg

= 258 nm = 258*10^-9 m

So,

K = (6.626*10^-34)^2/(2*(9.1*10^-31)*(258*10^-9)^2)

K = 3.624*10^-24 J

K = (3.624*10^-24)/(1.60*10^-19) eV

K = 2.26*10^-5 eV

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