SB257426
Very Important User
A question popped up in a physics textbook and I would just like to clarify my answer:
![Screen Shot 2023-01-24 at 7.51.17 pm.jpg](/data/attachments/37/37578-8640bc20bdc98f8985fc7381cf84b812.jpg)
a) My answer: Potential difference is positive so ΔV = ΔU/q
and since potential difference is positive so must be ΔU.
ΔU = q*ΔV = (-1.602 * 10^-19)(100)
= -1.602 * 10^-17
ΔU = -ΔKe
-1.602 * 10^-17 = -ΔKe
1.602 * 10^-17 = ΔKe
½mv^2 (f) - ½mv^2 (i) = 1.602 * 10^-17
½mv^2 (f) = 1.602 * 10^-17
then re-arranging for v and solving the values I got:
v = 1..86 * 10^7 m/s
b)
ΔU = q*ΔV = (-1.602 * 10^-19)(100)
= 1.602 * 10^-17
ΔU = -ΔKe
-1.602 * 10^-17 = -ΔKe
-1.602 * 10^-17 = ΔKe
½mv^2 (f) - ½mv^2 (i) = 1.602 * 10^-17
½mv^2 (f) = -1.602 * 10^-17
But if u rearrange for v then the expression inside the square root is negative ???????
c) I don't even know where to start, an alpha particle is 2 protons and 2 electrons???
Any help would be appreciated !!!
![Screen Shot 2023-01-24 at 7.51.17 pm.jpg](/data/attachments/37/37578-8640bc20bdc98f8985fc7381cf84b812.jpg)
a) My answer: Potential difference is positive so ΔV = ΔU/q
and since potential difference is positive so must be ΔU.
ΔU = q*ΔV = (-1.602 * 10^-19)(100)
= -1.602 * 10^-17
ΔU = -ΔKe
-1.602 * 10^-17 = -ΔKe
1.602 * 10^-17 = ΔKe
½mv^2 (f) - ½mv^2 (i) = 1.602 * 10^-17
½mv^2 (f) = 1.602 * 10^-17
then re-arranging for v and solving the values I got:
v = 1..86 * 10^7 m/s
b)
ΔU = q*ΔV = (-1.602 * 10^-19)(100)
= 1.602 * 10^-17
ΔU = -ΔKe
-1.602 * 10^-17 = -ΔKe
-1.602 * 10^-17 = ΔKe
½mv^2 (f) - ½mv^2 (i) = 1.602 * 10^-17
½mv^2 (f) = -1.602 * 10^-17
But if u rearrange for v then the expression inside the square root is negative ???????
c) I don't even know where to start, an alpha particle is 2 protons and 2 electrons???
Any help would be appreciated !!!