Titanium (6.94x10^-19) and silicon (7.77x10^-19) surfaces as irradiated with UV radiation with a wavelength of 231 nm. what is the wavelength of the electrons emitted by the titration surface?
What is a titration surface?
What are the numbers by Ti and Si?
7654
To determine the wavelength of the electrons emitted by the titanium surface due to the UV radiation, we can use the equation for the energy of a photon:
E = hc/λ
Where:
E represents the energy of the photon,
h is the Planck's constant (6.62607015 × 10^-34 J·s),
c is the speed of light (2.998 × 10^8 m/s), and
λ is the wavelength of the photon.
Since the UV radiation has a wavelength of 231 nm, we need to convert it to meters by multiplying by 10^-9:
λ = 231 nm x 10^-9
Now we can calculate the energy of the photons using the given wavelength:
E = (6.62607015 × 10^-34 J·s) x (2.998 × 10^8 m/s) / (231 nm x 10^-9)
E = (2.0028130 × 10^-24 J·m) / (2.31 x 10^-16 m)
E ≈ 0.867 × 10^-8 J
Next, we need to consider the work function (φ) of the titanium surface. The work function is defined as the minimum energy required to remove an electron from a material. In this case, we assume the energy of the photon (E) is equal to the work function (φ) of titanium:
φ = 0.867 × 10^-8 J
Finally, we can find the wavelength (λ') of the emitted electrons using the same equation as before:
λ' = hc/φ
λ' = (6.62607015 × 10^-34 J·s) x (2.998 × 10^8 m/s) / (0.867 × 10^-8 J)
λ' = (1.988 × 10^-24 J·m) / (0.867 × 10^-8 J)
λ' ≈ 2.29 × 10^-16 m
Therefore, the wavelength of the electrons emitted by the titanium surface due to the UV radiation is approximately 2.29 × 10^-16 meters.