Answer
a) $\lambda_c=1.32fm$
b) $E=1.51\times 10^{-10}J$
c) $E=mc^2=$ rest energy of the particle.
Work Step by Step
(a) We know that
$\lambda_c=\frac{h}{mc}$
We plug in the known values to obtain:
$\lambda_c=\frac{6.63\times 10^{-34}J.s}{(1.673\times 10^{-7}Kg)(3\times 10^8m/s)}$
$\lambda_c=1.32\times 10^{-15}m=1.32fm$
(b) As $E=\frac{hc}{\lambda_p}$
We plug in the known values to obtain:
$E=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{1.32\times 10^{-15}m}$
$E=1.51\times 10^{-10}J$
(c) We know that
$E=\frac{hc}{\lambda_p}$
$E=\frac{hc}{\lambda_c}$
$\implies E=\frac{hc}{\frac{h}{mc}}$
$E=mc^2=$ rest energy of the particle.
