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Ch.5 - Periodicity & Electronic Structure of Atoms

Chapter 5, Problem 130

One watt (W) is equal to 1 J/s. Assuming that 5.0% of the energy output of a 75 W light bulb is visible light and that the average wavelength of the light is 550 nm, how many photons are emitted by the light bulb each second?

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Hi everyone, This problem reads one watt is equal to one jewel per second. How many photons are released by a light bulb every second. If 4.5% of the energy produced by 55 watt light bulb is visible light and its average wavelength is nanometers. So the question that we want to answer here Is how many photons. Okay, so let's start off by writing out what 55 watts means. Okay, so we know we have 55 watts. And in the first sentence we're told one watt is equal to one jewel per second. So this 55 watts is equal to jewels per second. Okay. And we're told that every that we're told that 4.5% of the energy produced by 55 watt light bulb. So if we multiply our 55 watt light bulb, Okay, if we multiply this 55 watts or this jewels per second by 4.5%. So we're going to take 4.5 and divide it by 100. Okay, so we're gonna multiply it by 4.5%. This gives us 2.475 jewels per second. That represents that 4.5% of the energy. Okay, so now what we want to do is we want to write out the equation energy is equal to H. C. Over lambda. Because this equation, it shows the relationship between the energy of a photon and the wavelength of light. Okay, so here E represents energy H is Planck's constant C. Is the speed of light And lambda is the wavelength. Alright, so let's go ahead and calculate our energy and our energy is going to be in jewels per photon. Okay, so our final answer for energy is going to be jewels per photon. So let's go ahead and plug in our values. So we're going to get energy is equal to H. S. Plank's constant and that is 6. times 10 to the negative 34 C is our speed of light. And that value is three times 10 to the 8th. And our wavelength is in nanometers. Okay, but we want this in meters. So we're going to take 530 nanometers and we're going to multiply it. Bye. One times 10 to the negative nine m because that's the conversion from one nanometer to meter. Okay, so let's go ahead and do this calculation when we do the calculation, we're going to get 3. times 10 to the 19th times 10 to the negative 19th jewels per photon. Okay, but we just want photons and asked how many photons are released. So the answer we just got is jules per photon. So to get the number of photons. Okay, so we want number of photons. What we're going to do is we're going to take the 4.5% of the energy produced by this. 55 watt light bulb we calculated is 2.475 Jewels per second. And we're going to divide this by the energy we just calculated. So this is going to be divided by 3. times 10 to the negative 19 jewels per photon. Okay, so we'll see here that our our units of jewels cancel And we're going to be left with photon. Okay, so once we do this calculation, we're going to get 6.59, 9 Times 10 to the 18th Photon. So if we will go ahead and round this number for our final answer, we're going to get 6.6 times 10 to the 18 photon. This is going to be our final answer. All right. So this is how many photons are released by a light bulb every second. If 4.5% of the energy produced by a 55 light bulb is visible light and its average wavelength is 530 nanometers. That's it for this problem. I hope this was helpful
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Textbook Question
Orbital energies in single-electron atoms or ions, such as He+, can be described with an equation similar to the Balmer–Rydberg equation:

where Z is the atomic number. What wavelength of light in nanometers is emitted when the electron in He+ falls from n = 3 to n = 2?
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Imagine a universe in which the four quantum numbers can have the same possible values as in our universe except that the angular-momentum quantum number l can have integral values of 0, 1, 2...n + 1 (instead of 0, 1, 2..., n - 1). (a) How many elements would be in the first two rows of the periodic table in this universe?

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Imagine a universe in which the four quantum numbers can have the same possible values as in our universe except that the angular-momentum quantum number l can have integral values of 0, 1, 2...n + 1 (instead of 0, 1, 2..., n - 1). (c) Draw an orbital-filling diagram for the element with atomic number 12.

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Microwave ovens work by irradiating food with microwave radiation, which is absorbed and converted into heat. Assum-ing that radiation with l = 15.0 cm is used, that all the energy is converted to heat, and that 4.184 J is needed to raise the temperature of 1.00 g of water by 1.00 °C, how many photons are necessary to raise the temperature of a 350 mL cup of water from 20 °C to 95 °C?
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X rays with a wavelength of 1.54 * 10-10 m are produced when a copper metal target is bombarded with high-energy electrons that have been accelerated by a voltage difference of 30,000 V. The kinetic energy of the electrons equals the product of the voltage difference and the electronic charge in coulombs, where 1 volt-coulomb = 1 J. (a) What is the kinetic energy in joules and the de Broglie wavelength in meters of an electron that has been accel-erated by a voltage difference of 30,000 V?
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