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?

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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.