1B.15


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ShravanPatel2B
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Joined: Fri Aug 30, 2019 12:18 am

1B.15

Postby ShravanPatel2B » Sun Oct 13, 2019 6:00 pm

1B.15 The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 3 103 km?s21. (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 3 1016 Hz. How much energy is required to remove the electron from the metal surface? (c) What is the wavelength of the radiation that caused photoejection of the electron? (d) What kind of electromagnetic radiation was used?



I was able to solve through parts a and b without any difficulty but I was wondering if anyone could offer an explanation for part c. For parts a and b I got 2.0*10^-10 m and 1.66*10^-17 J

Does the originally given value of the velocity of the ejected electron get utilized in this part of the question?

Thanks!

Brian_Ho_2B
Posts: 221
Joined: Fri Aug 09, 2019 12:16 am

Re: 1B.15

Postby Brian_Ho_2B » Mon Oct 14, 2019 12:11 pm

ShravanPatel4G wrote:1B.15 The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 3 103 km?s21. (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 3 1016 Hz. How much energy is required to remove the electron from the metal surface? (c) What is the wavelength of the radiation that caused photoejection of the electron? (d) What kind of electromagnetic radiation was used?



I was able to solve through parts a and b without any difficulty but I was wondering if anyone could offer an explanation for part c. For parts a and b I got 2.0*10^-10 m and 1.66*10^-17 J

Does the originally given value of the velocity of the ejected electron get utilized in this part of the question?

Thanks!

Use the answer you found in part B to solve for the frequency of the photon (E = hv) and then use the frequency to solve for the wavelength of the photon (c = wavelength times frequency).

Ryan Chang 1C
Posts: 105
Joined: Sat Aug 24, 2019 12:17 am

Re: 1B.15

Postby Ryan Chang 1C » Mon Oct 14, 2019 2:12 pm

Yes, the original velocity of the ejected electron is used to solve this question.

The first step is to find the energy of the photon, which can be calculated by adding the kinetic energy of the ejected electron to the energy required to remove the electron from the metal surface. Recall that kinetic energy is equal to (1/2)mv^2.

E(photon) = 1.66x10^-17J + (1/2)(9.11x10^-31kg)(3.6x10^6m/s)^2 = 2.25x10^-17J

Next, you use the energy of the photon to find the wavelength using the equation E=(hc)/lambda, which is derived from E=hv and v=c/lambda

2.25x10^-17 = (6.626x10^-34J/s)(3x10^8m/s)/lambda, lambda = 8.8nm

So the answer is 8.8nm.

Hope this helps!


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