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

The recoil kinetic energy of the aluminum atom can be calculated using momentum conservation. This recoil energy is usually smaller than the kinetic energy of the photoelectron, which is primarily determined by the incident photon's energy.

Step-by-step explanation:

The recoil kinetic energy of an aluminum atom can be calculated using the momentum conservation principle. Since the photon is absorbed by the atom, the recoil momentum of the atom must be equal in magnitude and opposite in direction to the momentum of the photon. The recoil kinetic energy can be calculated using the equation KE = (p^2)/(2m), where p is the momentum and m is the mass of the atom.

It is important to note that the recoil energy of the atom is usually negligible compared to the kinetic energy of the photoelectron. This is because the mass of the atom is significantly larger than the mass of the electron, and therefore the recoil energy is much smaller. The photoelectron's kinetic energy is primarily determined by the energy of the incident photon, which depends on its wavelength or frequency.

Answer 2

`KE=3.529*10^(−27)\ J`

Step-by-step explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

`E=(hC)/(\lambda )`

Now by putting the values

`h=6.6* 10^(-34)\ m^2.kg/s`

`C=3* 10^(8)\ m/s`

`E=(6.6* 10^(-34)* 3* 10^(8))/(192* 10^(-9) )`

`E=1.03* 10^(-18) J`

We know that

Kinetic energy given as

`KE=(P^2)/(2m)`

`KE=(E^2)/(2mC^2)`

`KE=((1.03* 10^(-18))^2)/(2* 1.67* 10^(-27)(3* 10^8)^2)`

`KE=3.529*10^(−27)\ J`