Ch 25.1 Quiz

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Ch 25.1 Quiz

Post  Kathleen on Tue Apr 14, 2009 8:14 pm



Last edited by Kathleen on Thu Apr 16, 2009 10:04 am; edited 1 time in total

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question 3

Post  sjames on Wed Apr 15, 2009 7:51 am

I have been working on three for a while as well i cant figure it out does anyone know how to solve it? please help!!!!!

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question #1

Post  guest 00 on Wed Apr 15, 2009 10:53 am

the first equation f=c/lamda, i don't get why we have to do that if we r not using it...and i don't understand how you got to joules because if you multiplied the values to get E, 4.569e-19Js isn't the answer, so i dont really get it at all.

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question 1

Post  guest001 on Wed Apr 15, 2009 11:03 am

i finally figured #1 out..thanks kath..

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Question 3

Post  Guest 24 on Wed Apr 15, 2009 11:40 pm

The equation that is on sheet 7 reads
KEmax=e*Vs
we known that e is 1.6e-19 and we are given the stopping voltage(for me) is 6.437 hundred volts.
Since we are asked for the answer in eV and KEmax is normally in joules we can just cancel out the "e" term which is basically like a conversion factor and we just get KEmax=Vs
So to convert to just volts move the decimal place two spots over on the given number
e.g. 643.7 and that is your answer.
The answer is neg.

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question 1 and 4

Post  hwilson on Thu Apr 16, 2009 2:22 pm

I dont understand how to do question 1 or 4 has anyone figured them out?

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#1

Post  super mo on Thu Apr 16, 2009 4:13 pm

f=c/lambda (3*10^Cool/GIVEN
E=(h*f)/(1.6*10^-19) answer neg

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question 2 and 4

Post  john656 on Fri Apr 17, 2009 3:02 pm

Did someone one figure out how to do q 2 and 4 because for the life of me I cant!! please help?

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Maple TA 25_1

Post  User X on Fri Apr 17, 2009 9:35 pm

A photon has a wavelength of 5.131 x10-7m. What is its energy in eV (sheet 4) ? Indicate with a positive (negative) sign whether the photon energy increases (decreases) with increasing wavelength.

= (3E8)/(5.131E-7)
= 5.84681E14
THEN
=(5.84681E14)*(6.63E-34)
=(3.87644E-19)
THEN
=(3.87644E-19)/(1.6E-19)
=2.42277(NEGATIVE)
The reasoning (step by step):
1. Speed of light divided by the wavelength to get the frequency, and then multiply by "h" which is 6.63E-34 to get the amount of energy in terms of Joules, and then divide by 1.6E-19 in order to convert to eV.

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Maple TA 25_1

Post  User X on Fri Apr 17, 2009 9:53 pm

2. A "blackbody" (ideal emissivity - not essential here) emits an electromagnetic spectrum which has its maximum intensity at a wavelength of 587.1 nm. What is the temperature of the backbody in 0C (sheet 4') ? Indicate with a positive (negative) sign whether when heating up a piece of metal and observing the colors of the emitted light the blue (red) light appears first.

= (-2.9E-3)/(587.1E-9)
= -4939.53KELVIN
THEN
=(-4939.53)+(273)
= -4666.53CELSIUS

Reasoning:
The division step gets the temperature in terms of degrees kelvin, but the questions asks for the answer in celsius, so you need to add 273 to that number to get degrees celsius.

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Maple TA 25_1

Post  User X on Fri Apr 17, 2009 10:24 pm

I'm having some problems in understanding the concept behind question # 3. If someone could help me understand the rationale behind # 3 please.

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question 4

Post  cute611 on Sat Apr 18, 2009 10:01 pm

I have been working on question four but i cant quite figure it out has anyone figured out how to solve this problem yet??????????

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Maple TA 25_1

Post  User X on Sun Apr 19, 2009 12:55 am

4. Ok so number 4 is pretty straight forward, it's a simple step that you might have forgotten. So here's the deal:

4. The "workfunction" of a metal emitting electrons via the photoelectric effect is 1.817 eV. What is the threshold frequency i.e. the frequency below which no photoelectrons are emitted (sheet 9) ? Indicate with a positive (negative) sign whether for a certain metal an incident photon in the ultraviolet (infrared) is more likely to release an energetic electron.

= (6.63E-34)/(1.6E-19)
= (4.14375E-15)
THEN
=(1.817)/(4.14375E-15)
=(4.38492E14)

Reasoning: The reason you do the first step is to convert Planck's Constant from Joules into eV(electron-volts).

If someone can please explain the rationale behind question number 3 though, I would appreciate it.

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Maple TA 25_1

Post  User X on Sun Apr 19, 2009 8:13 pm

Anyone know how midterm 3 will be spilt up

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2.

Post  Helping on Mon Apr 20, 2009 4:14 pm

For #2 you have to subtract 273.15 from Kelvin to get Celcius!

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Re: Ch 25.1 Quiz

Post  DJ on Tue Apr 21, 2009 7:39 pm

User X wrote:Anyone know how midterm 3 will be spilt up

Midterm 3? You must mean final.

It'll probably be like last semester: 2/3 new material and the remainder review from MT's 1/2.

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