## Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

William Lan 2l
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### Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

How do you do questions 28, 29, and 30 in the post assessment module for the photoelectric effect?

Chem_Mod
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

Hi,

William Lan 2l
Posts: 73
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

28. Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1.
A. What is the kinetic energy of the ejected electron?

A. 3.01 x 10^25 J

B. 3.98 x 10^-19 J

C. 7.96 x 10^-19 J

D. 1.99 x 10^-19 J

E. None of the above

Equation should be 1/2mv^2. When I did it, I got 1.99e-16. I put none of the above as my answer, yet I still got it wrong. So how do you do this?

29 How much energy is required to remove an electron from one sodium atom?

A. 2.501 x 10^-22 J

B. 1.506 x 10^5 J

C. 2.501 x 10^-19 J

D. 9.069 x 10^28 J

E. None of the above

Well, it tells us the work function for sodium (150.6 kJ.mol-1). So, if it gives us the threshold energy (which is the energy required to remove an electron from one sodium atom, shouldn't the answer be 150.6 kJ-mol-1 (1.506 x 10^5 J)? I put B and I got it wrong.

Chem_Mod
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

For 28, your method is correct, however, I did get one of the above answers. Are you sure you input all the values into 1/2mv^2 correctly?
For 29, please take note of the units for the work function given. It is 150.6 kJ.mol-1, and the question is how much energy it takes to remove ONE SINGLE electron.

Rita Dang 3D
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

For 29, convert kJ to J first and then divide the number by the number of atoms per mole.

Varsha Sivaganesh 1A
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

For 28, what would you plug in for m in the equation E = 1/2mv^2? I am struggling with this one as well.

Sammy Thatipelli 1B
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

For m, the mass would be the mass of an electron. Therefore you would plus in 9.10939*10^-31 kg in the equation 1/2mv^2

Julia Meno 1D
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

Do you always plug in the mass of an electron as 9.10939 x 10-^31 kg for the kinetic energy equation 1/2mv^2?

Rita Dang 3D
Posts: 31
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

You always plug in 9.109x10^-31kg for m if you are calculating the kinetic energy of an electron.

Ashley Davis 1I
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

will we be given the mass of an e- and planck's constant in a problem or will we have to know them by heart?

Isabelle Bautista 3H
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

I did the same thing for 28 because I switched the mass of an electron to grams instead of keeping it in kilograms. I posted a question about this, but now I see that supposedly we are supposed to keep the mass of an electron in kg. It should work out once you do this.

Rakhi Ratanjee 1D
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

For 29, why do you divide by the number of atoms in a mole?

Naama 1A
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### Re: Photoelectric Effect: Post Module Assessment Q. 28, 29, and 30

You divide by the number of atoms in a mole because the question asks for the energy required to remove ONE electron, not a whole mole.