A metal with a threshold frequency of 2.15 x 10 15 s -1 emits an electron with a velocity of 7.03 x 10 6 m/s when radiation of 4.88 x 10 15 s -1 strikes the metal's surface. Calculate the mass of the electron.

In this practice question it says, a metal with a threshold frequency of 2.15 times 10 to the 15 seconds inverse emits an electron with a velocity of 7.03 times 10 to the 6 meters per second, when radiation of 4.88 times 10 to the 15 seconds inverse strikes the metal's surface. Calculate the mass of the electron. Alright. So we're talking about threshold frequency, we're talking about velocity here, and we're talking about radiation. First of all, remember that threshold frequency is another name for binding energy. And here they're giving us the frequency of energy, of radiation. Remember that's part of the electromagnetic spectrum. So this number here connects us to the energy of a photon. And remember, what formula connects binding energy and energy of a photon? Well, that would be the energy of a photon equals the energy, binding energy, plus the kinetic energy. This is the formula we're going to utilize to solve for the mass of our electron in this case. Now, here we're going to say, they're giving us the radiation, the frequency of it, we can use that to find the energy of our photon. We're going to say the energy of our photon equals Planck's constant times, well, here they're giving us seconds inverse, so that's frequency, so this would be times frequency. Planck's constant is 6.626 times 10 to the negative 34 joules times seconds. Plug in our frequency, 4.88 times 10 to the 15 seconds inverse. Seconds and seconds inverse cancel out. So we're gonna have joules here initially. So we're gonna have 3.233 times 10 to the negative 18 joules. Plug that in. Next, we have binding energy which is connected to threshold frequency. Binding energy can utilize the same formula. The difference now is we're still using Planck's constant but the frequency is different. The The threshold frequency is 2.15 times 10 to 15 seconds inverse. That's what we'll plug in for frequency. So same Planck's constant, plug in our frequency, Seconds and seconds inverse cancel out, so we're going to get 1.429x10-eighteen joules. So that goes here for binding energy. Now, kinetic energy. Remember the formula for kinetic energy equals half times the mass of our object, or in this case, electron in kilograms times our velocity squared. So we're gonna have half, we don't know the mass in this case of what it's giving us, but we don't know the velocity. 7.03 times 10 to the 6 meters per second, and that's going to be squared. We're going to subtract 1.429 times 10 to the negative 18 joules from both sides. Times 10 to the negative 18 joules. So here when we do that, we're going to get initially let's see. So we're going to get here 1.8089 times 10 to the negative 18 joules, and that equals half times my mass, which I still don't know. I'm going to square this right now. When I do that, I get 4.9421 times 10 to the 13 meters squared over seconds squared. Next, multiply both sides by 2. So when I do that, I'm gonna hit now. And remember, one more thing when we're doing our calculations, remember that 1 joule is equal to kilograms times meters squared over seconds squared. So I'm gonna do that here. So when I multiply this joules by 2, it's gonna give me 3.6178 times 10 to the negative 18. Again, 1 joule is equal to kilograms times meters squared over seconds squared. So let's substitute in those units. That's gonna equal now, mass times the velocity that I squared. Now divide both sides by 4.9421x10113 meters squared over second squared on both sides. So 4.9421 times 10 to the 13 meters squared over second squared. So what cancel out here? My meter squares cancel out, my second squares cancel out, so my mass at the end will be in kilograms. We're going to say here, m equals 7.32 times 10 to the negative 32 kilograms. Here we have our answer with 3 sig figs because all the values given to us within the question also have 3 sig figs. So this would be our final answer.

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9. Quantum Mechanics

Photoelectric Effect

General Chemistry

9. Quantum Mechanics

Photoelectric Effect

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Multiple Choice

A metal with a threshold frequency of 2.15 x 10¹⁵ s^-1 emits an electron with a velocity of 7.03 x 10⁶ m/s when radiation of 4.88 x 10¹⁵ s^-1 strikes the metal's surface. Calculate the mass of the electron.

6.97 x 10^-22 kg

4.98 x 10^-31 kg

2.96 x 10^-21 kg

7.32 x 10^-32 kg

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Verified step by step guidance

Identify the given values: threshold frequency (\( \nu_0 = 2.15 \times 10^{15} \text{s}^{-1} \)), frequency of incident radiation (\( \nu = 4.88 \times 10^{15} \text{s}^{-1} \)), and velocity of the emitted electron (\( v = 7.03 \times 10^{6} \text{m/s} \)).

Use the photoelectric effect equation to find the kinetic energy of the emitted electron: \( KE = h(\nu - \nu_0) \), where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{Js} \)).

Calculate the kinetic energy (\( KE \)) using the given frequencies: \( KE = 6.626 \times 10^{-34} \text{Js} \times (4.88 \times 10^{15} \text{s}^{-1} - 2.15 \times 10^{15} \text{s}^{-1}) \).

Relate the kinetic energy to the velocity of the electron using the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its velocity.

Solve for the mass of the electron (\( m \)) by rearranging the equation: \( m = \frac{2 \times KE}{v^2} \). Substitute the calculated kinetic energy and the given velocity to find the mass.