The singly-ionized helium atom: two protons plus two neutrons in the nucleus \"o

Question

The singly-ionized helium atom: two protons plus two neutrons in the nucleus "orbitied" by one electron at a constant distance of 2.00 x 10^-11 m. 1. what is the magnitude of the attractive electric force between the nucleus and the electron? 2. What is the magnitude of the electric field at the electrons location? 3.is its direction towards or away from the nucleus?

1. i used Coulomb's Law but unsure of if my values are correct?

F = k | q1q2| / r^2
   = (8.988 x 10^9 Nm^2/C^2) | (1.602 x 10^-19C)(1.602 x 10^-19C) | / (2 x 10^-11m)^2

   = 3.600 x 10^12 N   or do we just calc k with 1.602 x 10^-19C once? but then you can't cancel out the C with out it....

2. electric force is E = 1 / |q| * F The note in the queston says that you can answer 1 and 2 out of order. How can you calculate E without knowing Force? do we use electric field due to point charge? E = k * (|q|)/r^2?? then that would be solvable but unsure..

3. no idea

Thank you!

Explanation / Answer

1. Yes what u did is correct almost, u missed some minor things, first, u forgot the nucelus has two protons, so one of the charge is not 1.602 it's twice of it,

2. The idea you got for second part is also correct, we can approach in that way.

3.As the force between electron and proton is attractive, the direction of force is towards the nucelus.

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