Answer
(a) $7.0\times 10^{-25}Kg\frac{m}{s}$
(b) $K.E=2.7\times 10^{-19}J=1.7eV$
Work Step by Step
(a) According to Heisenberg's priniciple
$\Delta p\geq\frac{h}{2\pi}$
We plug in the known values to obtain:
$\Delta p\times 0.15\times 10^{-9}\geq1.055\times 10^{-34}$
$\implies \Delta p\geq7.0\times 10^{-25}Kg\frac{m}{s}$
(b) We can find the kinetic energy of the electron as
$K.E=\frac{p^2}{2m}$
We plug in the known values to obtain:
$K.E=\frac{(7.03\times 10^{-25})^2}{2\times 9.1\times 10^{-31}}$
$K.E=2.7\times 10^{-19}J=1.7eV$
