Question

A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron traveling in the opposite direction. Assume the recoil momentum is taken up by a single aluminum atom on the surface. Calculate the recoil kinetic energy of the atom. Would this recoil energy significantly affect the kinetic energy of the photoelectron?

Answer 1

Answer:

$KE=3.529\times10^{−27}\ J$

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

$E=\dfrac{hC}{\lambda }$

Now by putting the values

$h=6.6\times 10^{-34}\ m^2.kg/s$

$C=3\times 10^{8}\ m/s$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }$

$E=1.03\times 10^{-18} J$

We know that

Kinetic energy given as

$KE=\dfrac{P^2}{2m}$

$KE=\dfrac{E^2}{2mC^2}$

$KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}$

$KE=3.529\times10^{−27}\ J$