Answer
(a) $KE = 2.5\times 10^{-21}~J$
(b) The spring constant of the carbon-carbon bond is 2.0 N/m.
Work Step by Step
(a) We can find the kinetic energy of the carbon atom.
$KE = \frac{1}{2}mv^2$
$KE = \frac{1}{2}(2.0\times 10^{-26}~kg)(500~m/s)^2$
$KE = 2.5\times 10^{-21}~J$
(b) We can find the spring constant of the carbon-carbon bond.
$\frac{1}{2}kx^2 = U_s$
$k = \frac{2~U_s}{x^2}$
$k = \frac{(2)(2.5\times 10^{-21}~J)}{(0.050\times 10^{-9}~m)^2}$
$k = 2.0~N/m$
The spring constant of the carbon-carbon bond is 2.0 N/m.
