Answer
$K = 6.2~eV$
Work Step by Step
We can find the de Broglie wavelength:
$y = \frac{2\lambda d}{a}$
$\lambda = \frac{y~a}{2 d}$
$\lambda = \frac{(0.0033~m)(15\times 10^{-9})~m}{(2)(0.050~m)}$
$\lambda = 4.95\times 10^{-10}~m$
We can find the speed:
$v = \frac{h}{m \lambda}$
$v = \frac{6.626\times 10^{-34}~J~s}{(9.109\times 10^{-31}~kg) (4.95\times 10^{-10}~m)}$
$v = 1.47\times 10^6~m/s$
We can find the kinetic energy of each electron:
$K = \frac{1}{2}mv^2$
$K = (\frac{1}{2})(9.109\times 10^{-31}~kg)(1.47\times 10^6~m/s)^2$
$K = 9.84\times 10^{-19}~J$
$K = (9.84\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$K = 6.2~eV$
