A laser emits 1.0 × 10 19 photons per second from an excited state with energy E 2 = 1.17 eV. The lower energy level is E 1 = 0 eV. What is the wavelength of this laser?

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem with the fourth excited state at E subscript four equals 3.41 electron volts. And the third at E subscript three equals 2.11 electron volts. A laser discharges photons at a rate of 2.4 multiplied by 10 to the power of 20 per second find the wavelength associated with the light released by this laser system. So that's our end goals. We're trying to figure out what the wavelength will be that's associated with this light released by the specific laser system. Awesome. So we have some multiple choice answers. They're all in the same units of nanometers because we're trying to find a wavelength and the standard unit for wavelength is nanometers. So that's a good sign and that's good. So we, so let's read off our multiple choice answers to see what our final wavelength value might be. So A is 225 B is 841 C is 954 and D is 525. Awesome. So, first off, let us write down all of our known and unknown variables. We know that the energy level at the fourth excited state is equal to 3.41 electron volts. And this is the energy of the excited state. We also know the energy at the or the third excited state. The energy is 2.11 electron volts. And this is the energy of the lower state. And we do not know what lambda the wavelength is. And this is the wavelength associated with the light released by this particular laser system. So the energy difference delta E between the two states can be written as delta E is equal to E subscript four minus E subscript three which is equal to 3.41 electron volts subtract or minus 2.11 electron volts, which when we perform that subtraction, we will get 1.30 electron volts. Now, we can recall and use the equation for the energy of a photon which states that delta E is equal to Planck's constant multiplied by the speed of light all divided by the wavelength. So now we need to rearrange this equation to solve or LAMBDA. So we need to get lambda the wavelength all by itself. And when we do that, we will get that the wavelength lambda is equal to Planck's constant multiplied by the speed of light all divided by delta E. So at this stage, we can plug in our known variables to solve for lambda the wavelength. So the wavelength is equal to 4.135 multiplied by 10 to the power of negative 15 electron vaults. OK. And that is the numerical value for planks constant multiplied by the speed of light which the numerical value for that is 3.0 multiplied by 10 to the power of 8 m per second, all divided by our delta E which we've determined to be 1.30 ev electron volts. So when we plug that into a calculator, we will get 954 nanometers as our final answer. But if you're probably wondering like wait a minute, my calculator doesn't show 954 nanometers. So let's take a second here. So when you initially type this equation into your calculator, you probably got, I'll write it in blue. You probably got something along the lines of 9.5423 multiplied by 10 to the power of negative 7 m, which makes sense because our electron volts and you know, in seconds, we units will cancel out leaving us with just meters. So then we would have to convert meters to nanometers. So to do that, we multiply our meters value by 10 to the power of nine. And when we do that, we will get the 954.23 nanometers. But we need to round to the next whole number. So that will leave us with 954 nanometers. And that is our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter C which states that it's 954 nanometers. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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Problem 23a

Textbook Question

A laser emits 1.0 × 10¹⁹ photons per second from an excited state with energy E₂ = 1.17 eV. The lower energy level is E₁ = 0 eV. What is the wavelength of this laser?

Verified step by step guidance

Convert the energy difference between the two levels (E2 - E1) from electron volts (eV) to joules (J). Use the conversion factor: 1 eV = 1.602 × 10⁻¹⁹ J. The energy difference is ΔE = E2 - E1 = 1.17 eV.

Use the relationship between energy and wavelength given by the equation: E = h * c / λ, where E is the energy difference (in joules), h is Planck's constant (6.626 × 10⁻³⁴ J·s), c is the speed of light (3.00 × 10⁸ m/s), and λ is the wavelength (in meters). Rearrange the equation to solve for λ: λ = h * c / ΔE.

Substitute the values of h, c, and ΔE (converted to joules) into the equation to calculate the wavelength λ.

Convert the wavelength from meters to nanometers (1 m = 10⁹ nm) for a more convenient unit, if needed.

Verify the units and ensure the final wavelength is consistent with the expected range for laser emissions, typically in the visible or near-visible spectrum.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Photon Energy

The energy of a photon is directly related to its frequency and inversely related to its wavelength. It can be calculated using the equation E = hν, where E is the energy, h is Planck's constant (6.626 × 10^-34 J·s), and ν is the frequency. In this context, the energy difference between the excited state and the lower energy level determines the energy of the emitted photons.

Wavelength Calculation

The wavelength of a photon can be determined using the relationship between energy and wavelength, given by the equation λ = hc/E, where λ is the wavelength, h is Planck's constant, c is the speed of light (approximately 3.00 × 10^8 m/s), and E is the energy of the photon. This formula allows us to convert the energy of the emitted photons into their corresponding wavelength.

Laser Emission

A laser emits light through a process called stimulated emission, where photons stimulate excited atoms to emit additional photons of the same energy, phase, and direction. The number of photons emitted per second indicates the intensity of the laser, and in this case, it is essential to understand how the energy levels of the atoms relate to the characteristics of the emitted light.

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