1I ANS. 2.165*106ms-1
1. The problem statement, all variables and given/known data
The radius of the orbit of an electron in a hydrogen atom is 0.05 nm. Using the planetary model, calculate the orbital velocity needed by the electron to avoid being pulled into the nucleus by electrostatic attraction. Round up your answer to the nearest tenth.
2. Relevant equations
mv^2=k(e^2/r)
mass of electron -13.6?? (not sure)
v=velocity
k= coloumb constant
e=1.6x10^-19
3. The attempt at a solution
13.6(v^2)=k(q1q2/r2)(1.6x10^-19/5x10^11 m)
i think the equation is wrong :( im not sure about coloumbs constant and what it is..
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8 Answers
12. Relevant equations mv^2=k(e^2/r) That much is right. mass of electron -13.6?? (not sure) m nt sure abt d mass..
. When was the last time you put something on a scale and found that its weight is negative? :tongue: You should look up the electron mass. That's not too much to ask. v=velocity k= coloumb constant e=1.6x10^-19 OK 3. The attempt at a solution 13.6(v^2)=k(q1q2/r2)(1.6x10^-19/5x10^11 m) i think the equation is wrong :( im not sure about coloumbs constant and what it is.. m i rite...? plz check..
1no..i m nt sure ...dats specified na..i hve givn d doubts at d points
1answer r u sure abt it?
i am havng many doubts plz clarify dem...answer is a secondary thng
1itz simply
mv^2/r =kq^2/r^2
nw
r=radius (given)
m=9.1 *10^(-27) kg
v =nikalna hai :P
k=columb constnt =9*10^9
q=1.6 *10^-19 C
nikal do aab aap :P