Two positively charged ions of masses 16 a.u. and 17 a.u. carry one elementary charge each (1a.u.=1.6∗10−27kg,1e=1.6∗10−19C
). Initially with negligible speed, they are accelerated by a potential difference of 1000 V and then fly into a chamber with the magnetic field B = 0.1 T and hit the screen as shown, making half a circle. The whole setup is vacuumized so that there is no collisions of the ions with gas molecules.
a) What is the direction of the magnetic field? (Draw a schematics and indicate the field direction).
b) What is the distance ΔR between the two spots on the screen corresponding to the two ions?
a) The direction of the magnetic field should be into the page. This means that the ions are moving in a direction perpendicular to the magnetic field.
b) To find the distance ΔR between the two spots on the screen corresponding to the two ions, we can first calculate the radius of the circular motion each ion undergoes. The radius of the circular motion can be calculated using the formula:
r = (m*v)/(q*B)
where m is the mass of the ion, v is the velocity of the ion, q is the charge of the ion, and B is the magnetic field strength.
Since the ions are accelerated by a potential difference of 1000 V, we can calculate the velocity of the ions using the formula:
qV = 1/2*m*v^2
Plugging in the given values, we find the velocities of the two ions to be:
v1 = sqrt((2*1*1000*1.6*10^-19)/(16*1.66*10^-27)) = 2.94*10^5 m/s
v2 = sqrt((2*1*1000*1.6*10^-19)/(17*1.66*10^-27)) = 2.82*10^5 m/s
Now, we can calculate the radius of the circular motion for each ion:
r1 = (16*1.66*10^-27*2.94*10^5)/(1*0.1) = 4.74*10^-3 m
r2 = (17*1.66*10^-27*2.82*10^5)/(1*0.1) = 5.51*10^-3 m
The distance ΔR between the two spots on the screen can be found by taking the difference between the radii of the two circular motions:
ΔR = r2 - r1 = 5.51*10^-3 m - 4.74*10^-3 m = 0.77*10^-3 m = 0.77 mm
Therefore, the distance ΔR between the two spots on the screen corresponding to the two ions is 0.77 mm.