12. If alpha particles are caused to fall through a potential difference of 1500 volts, determine their final speed if they were initially at rest. Alpha particles are Helium nuclei composed of two protons and two neutrons.
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Ke= 1/2 m v^2= Voltate*charge
solve for v.
To determine the final speed of alpha particles falling through a potential difference of 1500 volts, we need to use the principles of energy conservation.
1. Start by recalling the relationship between electric potential difference, voltage (V), and electric potential energy (U) for a charged particle:
U = qV
Where U is the change in electric potential energy, q is the charge of the particle, and V is the voltage.
2. In this case, the alpha particles are not charged, so their electric potential energy remains constant.
U = qV = 0
This means that the change in electric potential energy is zero.
3. The change in electric potential energy is equal to the change in kinetic energy.
ΔK = ΔU
Since ΔU = 0, we have:
ΔK = 0
This means that the change in kinetic energy is zero.
4. Initially, the alpha particles are at rest, so their initial kinetic energy is zero:
K1 = 0
5. The final speed of the alpha particles can be calculated using the formula for kinetic energy:
K2 = 0.5 * m * v^2
Where K2 is the final kinetic energy, m is the mass of the alpha particles, and v is the final speed.
6. Setting K1 = K2, and substituting the given values, we have:
0 = 0.5 * m * v^2
Since m is a constant for alpha particles (approximately 6.64 x 10^-27 kg), we can solve for v:
0 = 0.5 * (6.64 x 10^-27 kg) * v^2
Rearranging the equation:
v^2 = 0
v = 0 m/s
Therefore, the final speed of the alpha particles, after falling through a potential difference of 1500 volts, will be 0 m/s.