What is the maximum velocity (in m/s) acquired by an alpha particle when accelerated between two plates with a potential difference of 580.0 volts?
Note: An alpha particle is essentially the nucleus of a helium atom - 2 protons and 2 neutrons.
To find the maximum velocity acquired by an alpha particle when accelerated between two plates with a potential difference of 580.0 volts, we can use the equation for the kinetic energy of a particle accelerated through a potential difference.
The kinetic energy (K) of a charged particle accelerated through a potential difference (V) is given by the equation:
K = qV
Where q is the charge of the particle.
Since an alpha particle has a charge of +2e (where e is the elementary charge), we can substitute q = +2e into the equation:
K = (+2e)V
Now, since kinetic energy is given by the equation:
K = 0.5mv^2
Where m is the mass of the particle and v is its velocity. We can rearrange the equation to solve for v:
v = sqrt((2K) / m)
The mass (m) of an alpha particle is approximately 6.64 x 10^-27 kg.
Now, we have the kinetic energy (K) from our first equation, which is (2e)V. Substituting the values:
v = sqrt((2(2e)V) / m)
Using the value of the elementary charge e = 1.6 x 10^-19 C and the potential difference V = 580.0 volts:
v = sqrt((2(2)(1.6 x 10^-19 C)(580.0 V)) / (6.64 x 10^-27 kg))
Calculating this expression will give us the maximum velocity (in m/s) acquired by the alpha particle when accelerated between the plates.