Question

What wavelength of light contains enough energy in a single photon to ionize a hydrogen atom?

Answer 1

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.)

Answer 2

Wavelength of the light is about 9.14 × 10⁻⁸ m

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Further explanation

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

$\large {\boxed {E = h \times f}}$

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

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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.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$

$\large {\boxed {E = qV + \Phi}}$

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 !

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Given:

energy of photon = E = 13.6 eV = 2.176 × 10⁻¹⁸ Joule

Unknown:

wavelength of light = λ = ?

Solution:

$E = h \times \frac{c}{\lambda}$

$2.176 \times 10^{-18} = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda}$

$2.176 \times 10^{-18} = 1.989 \times 10^{-25} \div \lambda$

$\lambda = (1.989 \times 10^{-25}) \div (2.176 \times 10^{-18})$

$\lambda \approx 9.14 \times 10^{-8} \texttt{ m}$

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Learn more

• Photoelectric Effect : brainly.com/question/1408276
• Statements about the Photoelectric Effect : brainly.com/question/9260704
• Rutherford model and Photoelecric Effect : brainly.com/question/1458544

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Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics

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Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light