Textbook Question 1B.15

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Evonne Hsu 1J
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Textbook Question 1B.15

Postby Evonne Hsu 1J » Tue Oct 20, 2020 12:00 pm

Can someone please explain how to do this question? Thank you!
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Sid Panda 2A
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Re: Textbook Question 1B.15

Postby Sid Panda 2A » Tue Oct 20, 2020 12:41 pm

Evonne Hsu 1J wrote:Can someone please explain how to do this question? Thank you!


To solve part a, I think you would have to use the DeBroigle Equation to solve for the wavelength of the electron, which is wavelengthe- = h/(me- * ve-).

For part b, you would have to use the equation E=h*v, and plug in 2.50*10^16 Hz for the frequency of the photon.

For part c, you would have to use the equation c=wavelength* v, and plug in 2.50*10^16 Hz for the frequency to solve for the wavelength.

For part d, depending on you answer that you get in part c, you would determine what type of radiation was used. I'm guessing it could be UV radiation, since is commonly used in a Photoelectric Effect experiment like this one. But, again, it depends on the answer you get in part c.

JonathanSung_2G
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Re: Textbook Question 1B.15

Postby JonathanSung_2G » Tue Oct 20, 2020 12:47 pm

a. Use the DeBroglie wavelength equation and plug in 9.11x10^-31kg for the mass of the electron. Make sure to convert the velocity of the electron to m/s by multiplying the velocity given by 1000. You should get 0.200nm as your wavelength.
b. This part of the question is essentially telling you that the threshold energy is reached when the frequency reaches 2.50x10^16Hz. Plug this value in to the equation E=hv and solve for E(threshold energy).
c. The energy of the radiation that caused the ejection of the electron is equal to the threshold energy added to the kinetic energy: E(photon)=E(threhold)+E(kinetic). You already know the value of the threshold energy. Use the value of the velocity given and the mass of the electron to find the kinetic energy. By adding the two together, you get the energy of the photon. Now, you can solve for the wavelength using lambda=hc/E(photon).
d. The previous problem gives you the wavelength in the range of x-ray electromagnetic radiation. Hope this helped!

Eliana Carney 2F
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Joined: Wed Sep 30, 2020 9:31 pm

Re: Textbook Question 1B.15

Postby Eliana Carney 2F » Sat Oct 24, 2020 12:39 pm

Hey Evonne!

To solve part A, you first find the electron's momentum using the formula: momentum(p) = (mass(m)) x (velocity(v)). Make sure to convert the given velocity from kg/s to m/s. Then, you use the momentum to find the energy of the electron with the formula: energy(E) = (momentum(p)) x (speed of light(c)). Once you find the energy of the electron, you use it to calculate the frequency using the formula: frequency = (energy(E)) / (Planck's constant(h)). After you calculate the frequency, you use this value to calculate the wavelength with the formula: wavelength = (speed of light(c)) / (frequency).

For part B, you calculate the energy using the formula: energy(E) = (Planck's constant(h)) x (frequency).

For part C, you must first remember that (energy of light) - (energy to remove electron) = excess/kinetic energy. You can rearrange this equation to find energy of light by writing it as energy of light = (excess/kinetic energy) - (energy to remove electron). Another equation to remember is the equation for kinetic energy: kinetic energy = (1/2) (mass(m))(velocity(v))^2. By combining these two equations you get energy of light = ((1/2) (mass(m))(velocity(v))^2) - (energy to remove electron). From here, you just plug in to find the energy of the light. Remember, when solving for kinetic energy you must convert the given velocity from kg/s to m/s. The energy to remove the electron is the value you calculated in part B. Once you calculate the energy, you then use it to calculate the frequency using the formula: frequency = (energy(E)) / (Planck's constant(h)). After you calculate the frequency, you use this value to calculate the wavelength with the formula: wavelength = (speed of light(c)) / (frequency).

For part D, you take the wavelength value you found in part C and compare it to an electromagnetic spectrum to see what kind of electromagnetic radiation was being used.

Hope this helped!


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