Calculating Charge on a 1.72 g Gold Nugget

  • Thread starter Thread starter Purduenub03
  • Start date Start date
  • Tags Tags
    Charge Gold
AI Thread Summary
To calculate the net charge of a 1.72 g gold nugget after removing 1% of its electrons, first determine the total number of electrons in the nugget, which is approximately 4.154 x 10^23. Removing 1% of these electrons results in about 4.154 x 10^21 electrons being removed. Each electron has a charge of -1.602 x 10^-19 C, so the total charge removed is approximately -6.65 x 10^2 C. The nugget's net charge becomes positive due to the loss of negatively charged electrons, resulting in a net positive charge. The calculation confirms the importance of accurately determining the number of electrons and applying the charge of an electron to find the net charge.
Purduenub03
Messages
11
Reaction score
0
I'm sure I am missing something simple here but...

Suppose a 1.72 g nugget of pure gold has zero net charge. What would be its net charge after it has 1% of its electrons removed?

So i know that an electron is 9.109 x 10^-31 kg, and has a charge of -1.602 x 10^-19C. However, how do i know the split of electrons and protons in the gold bar?

I'd assume I am trying to get to mass of electrons / mass of electron, that times the charge of an electron divided by 100?
 
Physics news on Phys.org
How many electrons in a Gold atom ? How many atoms in 1.72 g ?
 
79 electrons in a gold atom

1.72/196.96654 = .008732 atoms

so .008732 * .79 * 1.602 x 10 ^-19 ?
 
Okay,you're on the wrong track...

197g gold----------------------------->6.023 10^{23} atoms

1.72g gold----------------------------> x atoms

1 atom gold---------------------------> 79 electrons
x atoms gold---------------------------> y electrons


What do you get for "y"??

Daniel.
 
(1.71 / 197) * 6.023 X 10 ^23 * 79 = 4.154 x 10 ^ 23

AH!

and that divided by 100 x the charge is the answer

thank you

Ryan
 
Last edited:
Not so fast !

You've only found the number of electrons removed. Multiply this number by the charge on an electron to get the net positive charge.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top