Question

(a) An electron has a kinetic energy of 6.23 eV. Find its wavelength. nm (b) A photon has energy 6.23 eV. Find its wavelength. nm

Answer 1

a)KE = 6.23 eV = 9.98 x 10^-19 J

We know that, KE related to momentum as;

E = p^2/2m

P = sqrt (2 E m) = sqrt ( 2 x 9.98 x 10^-19 x 9.1 x 10^-31) = 1.348 x 10^-24 kg-m/s

Now the wavelength related to momentum as:

lambda = h/p = 6.626 x 10^-34 / 1.348 x 10^-24 = 4.92 x 10^-10 m = 0.492 nm

Hence, lambda = 0.492 nm

b)We know that,

E = hc/lambda

lambda = hc/E

lambda = 6.626 x 10^-34 x 3 x 10^8 / 9.98 x 10^-19 = 1.99 x 10^-7 m = 199 x 10^-9 m = 199 nm

Hence, lambda(photon) = 199 nm.