A proton moves in a circular path whose orbit is 14 cm in a...

A proton moves in a circular path whose orbit is 14 cm in a uniform field of 0.57 T perpendicular to the speed of the proton. Find the velocity of the proton. 7.6x10^6 m/s 2.9x10^6 m/s 4.7x10^6 m/s 9.1x10^6 m/s A gold foil achieves the photoelectric effect after passing the minimum frequency of 1.14x10^15 Hz. Now it is at a frequency of 3.67x10^15 Hz. Find the kinetic energy that these photons have at this point. a. 7.56x10^-19 J b. 2.86x10^-19 J c. 1.68x10^-19 J d. 4.21x10^-19 J

Transcript

0:00 Hi there.

00:01 So for the first problem in here, we are told that a proton, we know that the mass of a proton, well, the mass of a proton is equal to 1 .67 times 10 to minus 27 kilograms.

00:17 And it is moving in a circular path whose orbit radius is given.

00:26 That radius is equal to 14 centimeters.

00:29 We know that 14 centimeters in meters is 0 .14 meters, and is in an uniform field, magnetic field, that is equal to a value of 0 .57 tesla, perpendicular to the speed of the proton.

00:51 Now, with that said, we need to find the velocity of the proton.

00:55 So that's the question for the first part of this problem.

00:59 Now we know that the radius of the circular path of a charged particle that is inside a magnetic field is just simply the following.

01:12 It's the mass of the particle times its speed and this divided by the charge of the particle times the magnetic field.

01:24 So what we need to do is to solve for this speed so we will have that that is.

01:28 The part between the radius, the charge, and the magnetic field, and this divided by the mass of the particle.

01:34 And now we just need to simply substitute all of these values.

01:38 So we start with the radius that is 0 .14 meters, which is the same as 14 centimeters, times the charge of a proton.

01:46 The charge of a proton is the same charge of an electron, but with a positive sign, so that will be 1 .6 times 10 to the minus 19 columns.

01:55 These times the magnetic field which is 0 .57 tesla and this divided by the mass which we know for the mass of proton is 1 .67 times 10 to the minus 27 kilograms.

02:18 So by doing this product we obtain a speed off 7 .65 times 10 to the 6 meters per second.

02:33 So the current, well, let me see if we are given that option.

02:40 Well, that corresponds to the first option in here.

02:44 We can approximate this to 7 .6 times 10 to 6 meters per second.

02:50 So that's a solution for this part of the problem.

02:53 Now for the second problem in here, we are told that a gold foil shifts the photoelectric epit after passing the minimum frequency that is equal to a value of 1 .14 times 10 to the 15 hertz if i'm correct yes and now it is at a frequency of 3 .67...

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