There's probably a much quicker, easier way to do it, but I don't work with this stuff every day so this is the way I have to do it:
First, I searched the "ionization energy" of Hydrogen on Floogle. That's how much work it takes to rip the one electron away from its Hydrogen atom, and it's 13.6 eV (electron-volts).
In order to find the frequency/wavelength of a photon with that energy, I need the energy in units of Joules.
1 eV = 1.602 x 10⁻¹⁹ Joule (also from Floogle)
13.6 eV = 2.179 x 10⁻¹⁸ Joule
OK. Now we can use the popular well-known formula for the energy of a photon:
Energy = h · (frequency)
or Energy = h · (light speed/wavelength)
' h ' is Max Planck's konstant ... 6.626 × 10⁻³⁴ m²-kg / s
Wow ! The only thing we don't know in this equation is the wavelength, which is what we need to find. That's gonna be a piece-o'-cake now, because we know the energy, we know ' h ', and we know the speed of light.
Wavelength = h · c / energy
Wavelength =
(6.626 x 10⁻³⁴ m²-kg/sec) · (3 x 10⁸ m/s) / (2.179 x 10⁻¹⁸ joule)
Wavelength = 9.117 x 10⁻⁸ meter
That's 91.1 nanometers .
It's not visible light (visible is between about 390 to 780 nm), but it's not as short as I was expecting. I thought it was going to be an X-ray, but it's not that short. X-rays are defined as 0.1 to 10 nanometers. This result is in the short end of Ultra-violet.
(You have no idea how happy I am with this result. I figured it out exactly the way I showed you, and I never peeked. Then, AFTER I had my solution, I went to Floogle and searched to see what it really is, and whether I came out anywhere close. I found it in the article on the "Lyman Series". It says the wavelength of the energy released by an electron that falls in from infinity and settles in the n=1 energy level of Hydrogen is 91.175 nm ! This gives me a big hoo-hah for the day, and I'm going to bed now.)
Wavelength of the light is about 9.14 × 10⁻⁸ m
[tex]\texttt{ }[/tex]
The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :
[tex]\large {\boxed {E = h \times f}}[/tex]
E = Energi of A Photon ( Joule )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
f = Frequency of Eletromagnetic Wave ( Hz )
[tex]\texttt{ }[/tex]
The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
E = Energi of A Photon ( Joule )
m = Mass of an Electron ( kg )
v = Electron Release Speed ( m/s )
Ф = Work Function of Metal ( Joule )
q = Charge of an Electron ( Coulomb )
V = Stopping Potential ( Volt )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
energy of photon = E = 13.6 eV = 2.176 × 10⁻¹⁸ Joule
Unknown:
wavelength of light = λ = ?
Solution:
[tex]E = h \times \frac{c}{\lambda}[/tex]
[tex]2.176 \times 10^{-18} = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda}[/tex]
[tex]2.176 \times 10^{-18} = 1.989 \times 10^{-25} \div \lambda[/tex]
[tex]\lambda = (1.989 \times 10^{-25}) \div (2.176 \times 10^{-18})[/tex]
[tex]\lambda \approx 9.14 \times 10^{-8} \texttt{ m}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Quantum Physics
[tex]\texttt{ }[/tex]
Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light
