The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 5.48 × 1014 s–1.
With what kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of λ = 225 nm?
For the first half I got 3.63*10^(-19)
and that was correct. I just can't get the second part.
how did u do the first part???
Scroll down to mathematical description.
http://en.wikipedia.org/wiki/Photoelectric_effect
8.83410667x10^-19
To find the kinetic energy of the ejected electrons when the metal is exposed to light with a certain wavelength, you need to use the equation for the energy of a photon.
The energy of a photon (E) is given by the equation:
E = h * ν
Where:
E is the energy of the photon
h is Planck's constant (6.626 x 10^-34 J·s)
ν is the frequency of the light
In this case, you have the wavelength (λ) of the light instead of the frequency. The relationship between wavelength and frequency is given by:
c = λ * ν
Where:
c is the speed of light (3.00 x 10^8 m/s)
You can rearrange this equation to solve for the frequency:
ν = c / λ
Now that you have the frequency (ν), you can use it to calculate the energy of the photon (E) using the equation above. Once you have the energy of the photon, you can calculate the kinetic energy of the ejected electrons using the equation:
KE = E - E0
Where:
KE is the kinetic energy of the ejected electrons
E is the energy of the photon
E0 is the energy required to eject an electron (threshold energy)
In this case, you already have the value of the threshold frequency (5.48 x 10^14 s^(-1)), which corresponds to the threshold energy (minimum energy to eject an electron).
Let's plug in the values and calculate the kinetic energy:
1. Calculate the frequency (ν) using the wavelength (λ):
ν = c / λ
ν = (3.00 x 10^8 m/s) / (225 x 10^(-9) m)
ν ≈ 1.33 x 10^15 s^(-1)
2. Calculate the energy of the photon (E):
E = h * ν
E = (6.626 x 10^-34 J·s) * (1.33 x 10^15 s^(-1))
E ≈ 8.80 x 10^(-19) J
3. Calculate the kinetic energy of the ejected electrons (KE):
KE = E - E0
KE = (8.80 x 10^(-19) J) - (3.63 x 10^(-19) J)
KE ≈ 5.17 x 10^(-19) J
Therefore, the kinetic energy of the ejected electrons when the metal is exposed to light with a wavelength of 225 nm is approximately 5.17 x 10^(-19) J.