Answer
a) $0.27nm$
b) $0.27nm$
Work Step by Step
(a) We know that
$\lambda_e=\frac{h}{P_e}$
$\implies \lambda=\frac{6.63\times 10^{-34}}{(9.11\times 10^{-31})(2.7\times 10^6)}$
$\implies \lambda_e=0.269\times 10^{-9}m=0.27nm$
(b) We can find the de Broglie wavelength of the photon as
$\lambda_p=\frac{h}{P_p}$
$\implies \lambda=\frac{6.63\times 10^{-34}}{(9.11\times 10^{-31})(2.7\times 10^6)}$
$\implies \lambda_p=0.269\times 10^{-9}m=0.27nm$
