## 1B.15 part b and c.

$\lambda=\frac{h}{p}$

Anna_Mohling_1D
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Joined: Wed Sep 30, 2020 9:56 pm

### 1B.15 part b and c.

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 2.50×10^16 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 am slightly confused about this question and am struggling with understanding the different types of energy and how they relate to equations. For part b, why don't we solve for the work function since it's asking how much energy is required to remove the electron? Also, for part c when solving for Ek, why do we add the Joules found in part b. to 1/2 mass(electron)v^2?

Abril Guanes 2A
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### Re: 1B.15 part b and c.

Hello! I'll break down what I did for each part.

A) For part a I used the wavelength = h/p equation. For the top part, h is just Planck's constant and p is momentum so (mass x velocity, which is given). So, it would look something like this wavelength = (6.626x10^-34)/(3.6x10^6)*(9.11x10^-31), which would give you the wavelength.

B) To find the energy required to remove the electron, E = h*v, and the frequency is given in the problem. This gives you the work function.

C) For this part, you would use the equation E(photon) = kinetic energy of the electron (1/2*mass of electron*(v)^2)+ the work function, which we found in part B: E(photon) = 1/2(9.11x10^-31)(3.6x10^6)^2 + (1.66x10^-17). That will give you the energy of the photon, and from there, you can use E = h*c/wavelength to find the wavelength. You add the answer from part B to the equation because the energy of the photon is equal to the energy required to remove the electron PLUS the excess energy of the electron, which is the velocity that was stated and we converted it to a wavelength.

D) Using the wavelength from C, you can determine the radiation based o where it falls in the spectrum.

I hope this makes sense!

Crystal Pan 2G
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### Re: 1B.15 part b and c.

For 1B.15 part b, you use the $E=h\nu$ which is $E=(6.626x10^{-34}J.s)(2.50x10^{16}s^{-1}) = 1.66x10^{-17}J$, and since no electrons are emitted until frequency reaches $2.50x10^{16}s^{-1}$, we can assume that is the work function.

For part c, you use $E(photon)= work function + Ek$, so $E= 1.66x10^{-17}J+(\frac{1}{2}(9.11x10^{-31}kg)(3.6x10^{6}m.s^{-1})^{2})$
and you get your answer $2.25x10^{-17}J$. From there, you calculate the wavelength using $wavelength=\frac{hc}{E}$ to get your wavelength.

Vivian Hoang 1H
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Joined: Wed Sep 30, 2020 10:09 pm

### Re: 1B.15 part b and c.

Crystal Pan 1B wrote:For 1B.15 part b, you use the $E=h\nu$ which is $E=(6.626x10^{-34}J.s)(2.50x10^{16}s^{-1}) = 1.66x10^{-17}J$, and since no electrons are emitted until frequency reaches $2.50x10^{16}s^{-1}$, we can assume that is the work function.

For part c, you use $E(photon)= work function + Ek$, so $E= 1.66x10^{-17}J+(\frac{1}{2}(9.11x10^{-31}kg)(3.6x10^{6}m.s^{-1})^{2})$
and you get your answer $2.25x10^{-17}J$. From there, you calculate the wavelength using $wavelength=\frac{hc}{E}$ to get your wavelength.

I was just about to ask the same exact question, but your explanation cleared it up! Thank you.