1B #15

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Natallie K 3B
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1B #15

Postby Natallie K 3B » Mon Oct 19, 2020 4:01 pm

1B.15 The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 × 10^3 km⋅ s^−1
(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 × 10^16Hz. 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 part A and B, but I am confused on the rest. Which E value is part C referring to?

Chem_Mod
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Re: 1B #15

Postby Chem_Mod » Mon Oct 19, 2020 4:04 pm

For part c, the photon needs enough energy to eject the electron and cause it to move with the given velocity. Therefore, the energy required is the work function plus the kinetic energy. The work function was found in part b, and 1/2mv^2 can be calculated with the given velocity and mass of an electron.

Jordan Tatang 3L
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Re: 1B #15

Postby Jordan Tatang 3L » Mon Oct 19, 2020 4:06 pm

Hi! So the energy that part c is referring to is the energy of the photon. Which is calculated by
E(photon) - work function = Kinetic energy

The work function is given to you in part b. You can find the kinetic energy of the electrons emitted in part a using the equation

kinetic energy = 1/2 mass (velocity)^2

By adding the work function to the kinetic energy you can find the energy per photon.

Hopefully that answers your question!

Natallie K 3B
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Re: 1B #15

Postby Natallie K 3B » Mon Oct 19, 2020 4:15 pm

Oh ok that makes sense! Just making sure, for part b do you use E=hv to find the answer?

Emmeline Phu 1G
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Re: 1B #15

Postby Emmeline Phu 1G » Mon Oct 19, 2020 4:32 pm

Natallie K 3B wrote:Oh ok that makes sense! Just making sure, for part b do you use E=hv to find the answer?

Hi! Yes, in order to solve for part B the equations E=h(nu) is used. Given the frequency for the electromagnetic radiation (2.50 × 10^16Hz), you're able to plug it into the equation (for the variable frequency) along with Planck's constant in order to find the energy required to remove an electron from the metal surface. Hope this helps! :)


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