I only need help with part b please!! explanation would be great Consider a prot
ID: 2199046 • Letter: I
Question
I only need help with part b please!! explanation would be great
Consider a proton, a deuteron (nucleus of deuterium, i.e., Hydrogen-2), and an alpha particle (nucleus of Helium-4), all with the same speed. These particles enter a region of uniform magnetic field B, traveling perpendicular to B.
a) What is the ratio of the deuteron's orbital radius to the proton's orbital radius? Correct answer: 2
b) What is the ratio of the alpha particle's orbital radius to the proton's orbital radius? Correct answer: 2
c) For the next two questions, consider the situation in which all three particles have the same kinetic energy. What is the ratio of the deuteron's orbital radius to the proton's orbital radius? Correct answer: sqrt2
d) What is the ratio of the alpha particle's orbital radius to the proton's orbital radius?
choices are 1/2, 1/sqrt2, 1, sqrt2, 2
Explanation / Answer
Centripetal force = mv^2 / r = magnetic force = qvB Solve for the radius of curvature: r = mv / qB The speed and magnetic field are held constant in this problem. The radius is proportional to the mass-to-charge ratio m/q. A) The deuteron has double the mass and the same charge as the proton. What does that do to the m/q ratio (and hence to r) B) The alpha has double the charge and 4x the mass. Same question. C) For a non relativistic particle, the kinetic energy is: T = 1/2 mv^2 So the speed is: v = sqrt (2T/m) plug that into the formula for the radius: r = sqrt (2Tm) / qB So if T and B are held constant, the radius is now proportional to the square root of mass and inversely proportional to charge. The deuteron has double the charge and the same mass, so... D) etc