Using the ionization energy chart, how much energy will it take to remove only 1 electron on the surface of Nickel? If you hit the surface of Nickel with a photo that has a wavelength of 90nm, what will the wavelength of that ejected electron be?
Do you not have the chart? Can you look up the energy needed to remove one electron? Then plug into the equation I wrote below.
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To find the energy required to remove one electron from the surface of Nickel, we need to refer to the ionization energy chart. You can find the ionization energy of Nickel by looking up its atomic number or using a periodic table. The ionization energy represents the amount of energy required to remove an electron from an atom in its ground state.
The first ionization energy (IE1) of Nickel is 7.64 electron volts (eV), which is equivalent to 1.23 x 10^-18 Joules (J). This means that to remove one electron from the surface of Nickel, it would require an energy of 1.23 x 10^-18 J.
Now, let's move on to the second part of your question regarding the wavelength of the ejected electron when the surface of Nickel is hit by a photo with a wavelength of 90nm.
To determine the wavelength of the ejected electron, we can use the principle of conservation of energy. The energy of the ejected electron will be equal to the energy of the incoming photon minus the ionization energy.
First, we need to convert the wavelength of the incoming photon from nanometers (nm) to joules (J). Since energy (E) is inversely proportional to wavelength (λ), we can use the equation:
E = hc/λ
Where:
E = energy of the photon
h = Planck's constant (6.626 x 10^-34 J·s)
c = speed of light in a vacuum (3.0 x 10^8 m/s)
λ = wavelength of the photon
Plugging in the values:
E = (6.626 x 10^-34 J·s * 3.0 x 10^8 m/s) / (90 x 10^-9 m)
E ≈ 2.20 x 10^-18 J
Next, we subtract the ionization energy from the energy of the photon to find the energy of the ejected electron:
E_ejected = E - IE1
E_ejected ≈ (2.20 x 10^-18 J) - (1.23 x 10^-18 J)
E_ejected ≈ 0.97 x 10^-18 J
Finally, to find the corresponding wavelength of the ejected electron, we can use the same equation as before:
λ_ejected = hc/E_ejected
Plugging in the values:
λ_ejected = (6.626 x 10^-34 J·s * 3.0 x 10^8 m/s) / (0.97 x 10^-18 J)
λ_ejected ≈ 2.18 x 10^-6 m or 2.18 μm (micro meters)
So, when the surface of Nickel is hit with a photon of 90nm wavelength, the ejected electron will have a wavelength of approximately 2.18 micrometers.