He' ions with a charge +e move at 1.00 x 10 m/s in a straight line through a velocity selector. Suppose the He' ions are replaced with He ions in which both electrons have been removed from the helium atom and the ion charge is +2e. At what speed must the He2+ ions travel through the same velocity selector in order to move in a straight line?
The velocity selector is designed to only allow particles with a specific velocity to pass through it without deflection. This velocity is given by:
v = E/B
where v is the velocity of the particle, E is the electric field in the velocity selector, and B is the magnetic field in the velocity selector.
Since the He' ion has a charge of +e, we can use the above equation to find the velocity of the He' ion:
v[He'] = E/B = (1.00 x 10^6 V/m)/(0.500 T) = 2.00 x 10^6 m/s
Now, we need to find the velocity of the He2+ ion that will allow it to travel in a straight line through the same velocity selector. Since the charge of He2+ is +2e, the force acting on it in the electric field will be twice as large as that on the He' ion. Therefore, we need to double the electric field strength in the velocity selector to maintain the same force-to-mass ratio for the He2+ ion:
E[He2+] = 2E = 2(1.00 x 10^6 V/m) = 2.00 x 10^6 V/m
Now we can use the equation for the velocity selector to find the velocity of the He2+ ion:
v[He2+] = E[He2+]/B = (2.00 x 10^6 V/m)/(0.500 T) = 4.00 x 10^6 m/s
Therefore, the He2+ ions must travel at a velocity of 4.00 x 10^6 m/s in order to move in a straight line through the same velocity selector.
To determine the speed at which the He2+ ions must travel through the same velocity selector in order to move in a straight line, we need to consider the forces acting on the ions.
In a velocity selector, the magnetic force experienced by a charged particle moving through a magnetic field is equal to the electric force experienced by the particle due to an electric field, which allows the particle to move in a straight line.
Given:
Charge on He' ions, q1 = +e
Velocity of He' ions, v1 = 1.00 x 10 m/s
Charge on He2+ ions, q2 = +2e
The magnetic force experienced by a charged particle moving through a magnetic field is given by the equation:
F_mag = q1 * v1 * B
The electric force experienced by the particle due to an electric field is given by the equation:
F_elec = q2 * E
Here, B represents the magnetic field strength, and E represents the electric field strength. Since the particles are moving in a straight line, the forces must be equal:
F_mag = F_elec
q1 * v1 * B = q2 * E
Substituting the values:
e * (1.00 x 10 m/s) * B = 2e * E
Simplifying the equation, we find:
B = 2E / (1.00 x 10 m/s)
Now, the speed of the He2+ ions through the same velocity selector can be found by considering the magnetic field and the charge/mass ratio of the He2+ ions.
The ratio of the charge to mass (q/m) for He2+ ions is twice that of He' ions, and the magnetic field required to move the He2+ ions in a straight line will also be twice as strong.
Therefore, the speed of the He2+ ions through the velocity selector will be the same as that of He' ions, i.e., 1.00 x 10 m/s.