Question

. An electron is in a box 2 nm across. What will be the lowest energy for the
electron

Answer 1

E = 0.094 eV

$E = n^{2} [\frac{h^{2} }{8mL^{2} } ]$

since its the lowest level n = 1

h is planck's constant $h = 6.626 * 10^{-34} J$

m is the mass of an electron $m = 1.109 * 10^{-31} kg$

L is length of the box $L = 2*10^{-9} m$

the answer will come out in Jules to get it in eV we will divide by the charge of an electron $e = 1.602 *10^{-19}$