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uranmaximum [27]
3 years ago
7

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?
Physics
1 answer:
alex41 [277]3 years ago
5 0

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

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