# Electromagnetism: Velocity of a subatomic particle (1 Viewer)

#### SB257426

##### New Member
A question popped up in a physics textbook and I would just like to clarify my answer:

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 !!!

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would the following working help?

Part (a):

Part (b):

Part (c):

$\bg_white m=6.64\times10^{-27}kg,\:q=+3.2\times10^{-1}$

$\bg_white v=\sqrt{\frac{2q\Delta V}{m}}$

$\bg_white v=\sqrt{\frac{2(3.2\times10^{-19}C)(1000V)}{6.64\times10^{-27}kg}}$

$\bg_white v=3.1\times10^5\:ms^{-1}$

#### wizzkids

##### Active Member
The equations you used are the correct equations, but, you need to recognise the question refers to speed, and not velocity.
Speed is always a positive number. Speed doesn't have a direction, it is not a signed quantity.
Kinetic energy is also a scalar quantity, so it doesn't have a direction.
OK, so now we are going to need the masses of these particles to substitute into the kinetic energy equation. You got the electron and the proton masses correct, The alpha particle is a positively charged helium nucleus, consisting of two protons and two neutrons.
I think you can finish the question now with these hints.
One last thing you need to consider.
If this was a real question in the HSC, you need to test whether any of the particles are reaching relativistic velocity. That means, does the speed exceed about 10% of the speed of light, which is 3 x 108 m/s (c). The particle we need to examine more closely is the electron, because it has such a small mass.
The speed you got for the electron is 1.86 x 107 m/s which is less than 10% of c, so you don't need to use the more complicated relativistic equation for mass and energy.

#### SB257426

##### New Member
The equations you used are the correct equations, but, you need to recognise the question refers to speed, and not velocity.
Speed is always a positive number. Speed doesn't have a direction, it is not a signed quantity.
Kinetic energy is also a scalar quantity, so it doesn't have a direction.
OK, so now we are going to need the masses of these particles to substitute into the kinetic energy equation. You got the electron and the proton masses correct, The alpha particle is a positively charged helium nucleus, consisting of two protons and two neutrons.
I think you can finish the question now with these hints.
One last thing you need to consider.
If this was a real question in the HSC, you need to test whether any of the particles are reaching relativistic velocity. That means, does the speed exceed about 10% of the speed of light, which is 3 x 108 m/s (c). The particle we need to examine more closely is the electron, because it has such a small mass.
The speed you got for the electron is 1.86 x 107 m/s which is less than 10% of c, so you don't need to use the more complicated relativistic equation for mass and energy.

Indeed, the textbook question does mention something about relativistic velocity several times through these types of questions, under some of the examples it even said that the speed was unrealistic so we would need to make a reference to the relativisitic velocity instead

#### SB257426

##### New Member
Would the following working help?

Part (a):

View attachment 37578

Part (b):

View attachment 37577

Part (c):

$\bg_white m=6.64\times10^{-27}kg,\:q=+3.2\times10^{-1}$

$\bg_white v=\sqrt{\frac{2q\Delta V}{m}}$

$\bg_white v=\sqrt{\frac{2(3.2\times10^{-19}C)(1000V)}{6.64\times10^{-27}kg}}$

$\bg_white v=3.1\times10^5\:ms^{-1}$
Thank you so much for your response !!!, it is very clear and logical