CHAPTER 1 QUESTION 33!!!


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Sheel Shah 1H
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CHAPTER 1 QUESTION 33!!!

Postby Sheel Shah 1H » Tue Oct 17, 2017 7:47 pm

Q: The velocity of an electron that is emitted from a metallic surface by a photon is 3.6x10^3 km/s.
a) What is the wavelength of the ejected electron.

I get the correct answer, 2.02x10^-10m, when I use de Broglie's equation; however, I cannot understand why we can't get the same answer if we calculate Ek=Ephoton (ejected)=0.5mv^2, and then use the E=hv equation to find the frequency, and then convert frequency to wavelength. It makes no sense since the two answers should be the same!?

K Stefanescu 2I
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Re: CHAPTER 1 QUESTION 33!!!

Postby K Stefanescu 2I » Tue Oct 17, 2017 8:16 pm

E=hv only applies to photons, since it uses the definition of v that is consistent with vw=c (with w=wavelength, v=frequency, and c=speed of light). This equation does not apply to electrons.

E=0.5mV^2 applies to particles that have mass, such as electrons. It is the same with deBroglie's equation, since it relies on p (momentum, which is p=mV). So you can only use deBroglie's equation on particles that have mass, such as electrons.

Riya Pathare 2E
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Re: CHAPTER 1 QUESTION 33!!!

Postby Riya Pathare 2E » Tue Oct 17, 2017 8:19 pm

The photon that causes the electron to be ejected off does not have the same wavelength. So if you are calculating using E = hv you are calculating the wavelength for the incoming photon not the ejected electron, whereas when you use De Broglie's equation you are calculating the wavelength of the ejected electron.

Sheel Shah 1H
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Re: CHAPTER 1 QUESTION 33!!!

Postby Sheel Shah 1H » Tue Oct 17, 2017 8:20 pm

but dont photons have mass as well?

Sheel Shah 1H
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Re: CHAPTER 1 QUESTION 33!!!

Postby Sheel Shah 1H » Tue Oct 17, 2017 8:23 pm

Riya Pathare 1D wrote:The photon that causes the electron to be ejected off does not have the same wavelength. So if you are calculating using E = hv you are calculating the wavelength for the incoming photon not the ejected electron, whereas when you use De Broglie's equation you are calculating the wavelength of the ejected electron.


But I am using the velocity given for the emitted photon and subbing it into so=1/2mv^2. I am then saying that ek=ephoton ejected, and using that to calculate wavelength! I am not calculating the wavelength of the incoming photon.

Alissa Stanley 3G
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Re: CHAPTER 1 QUESTION 33!!!

Postby Alissa Stanley 3G » Wed Oct 18, 2017 12:15 am

I also have a question on #33, but not related to what is above. I am wondering how, in part a, you get the mass? The answer in the book says that the mass is 9.1*10^-31, but that is not given in the question and I am unsure how you would figure that out.

Ethan-Van To Dis2L
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Re: CHAPTER 1 QUESTION 33!!!

Postby Ethan-Van To Dis2L » Wed Oct 18, 2017 12:32 am

The masses of electrons, protons, neutrons and all the other constants you need are in the very in the back of the book.

mhuang 1E
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Re: CHAPTER 1 QUESTION 33!!!

Postby mhuang 1E » Wed Oct 18, 2017 12:52 am

Alissa Stanley 3G wrote:I also have a question on #33, but not related to what is above. I am wondering how, in part a, you get the mass? The answer in the book says that the mass is 9.1*10^-31, but that is not given in the question and I am unsure how you would figure that out.


The mass of the electron is given in the back of the book, which is 9.10938 x 10^-31 kg. On later problems when needed to use the masses of the neutron and proton, it's also given in the same section.

Minie 1G
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Re: CHAPTER 1 QUESTION 33!!!

Postby Minie 1G » Fri Oct 20, 2017 12:36 am

Back to your original question, I understand it like this.
You cannot use the Ek = 0.5mv^2=hc/lambda because we are calculating the wavelength for an electron, and there is a given velocity (3.6x10^3 km/s, in the problem) for this. Electrons don't travel at the speed of light (the constant c, in the equation) and thus we must use the de Broglie equation instead. We use this Ek equation for photons because they are light particles and therefore do travel at the speed of light. That's why you get the wrong answer, because by using this formula instead of de Broglie's you're operating under the assumption that the electron is light.


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