(8c28p20) In a nuclear experiment a proton with kinetic...

(8c28p20) In a nuclear experiment a proton with kinetic energy 4.0 MeV moves in a circular path in a uniform magnetic field. If the magnetic field is B = 0.4 T what is the radius of the orbit? 10 MeV/T Submit Answer Incompatible units. No conversion found between \"MeV/T\" and the required units. Tries 0/4 Previous Tries What energy must an alpha particle (q = +2e, m = 4.0 u) and a deuteron (q = +e, m = 2.0 u) have if they are to circulate in the same orbit? Energy of the alpha? (MeV) Submit Answer Tries 0/4 Energy of the deuteron?

Transcript

00:01 Hello students here it is given that a proton having charge plus e and here a deutron having charge plus e and mass is mass of the deutron is given as atomic units and here alpha particle is given the charge is plus 2e and mass of the alpha particle is given as four units first we need to calculate the ratio of kinetic energies of their relation we know that kinetic energy of proton is equals to half of the kinetic energy of alpha particle because alpha particle's charge is directly proportional to 2 into protons charge of protons is equal to is equals to now therefore kb by k alpha is equal to 1 by 2 now we can calculate second one which is given as the ratio of here kinetic energy of teutron is also half of the kinetic energy of alpha particle therefore kd is given as kd by k alpha is given as kd is given as 1 by 2 and another case is that the proton kinetic energy and neutron kinetic energy is equals to 1.

01:58 Therefore, from these relations, the kinetic energy expression can be given as here, kp by k alpha equals to 1 by 2, kd by k alpha equals to 1 by 2, kd by k alpha equals to 1 by 2.

02:22 Now we can calculate the radius formula for the radius of deutron.

02:29 The ratio of radius of deutron and ratio of radius of proton is given as rd for fourth question.

02:46 Third question rd, this is third by rp.

03:01 Radice of deutron to the radius of proton is given as mass of the ratio of masses square root of m .d.

03:09 Divided by mp into kinetic energies kd divided by kp into their opposite charges.

03:22 Here kd therefore it is qp divided by here kp it is qd.

03:30 Now substituting the values we will get here mass of the neutron is equals to and mass of the proton is 1 and kd by kp we have to.

03:45 Calculated which is equal to kb by kd equal to 1 into here qp by qd e by e plus c by c this is the expression for it now it will get that r d by rp is equal to root 2 into 1 is equal to root 2 therefore radius of deutron is equal to root 2 into radius of proton.

04:27 It is given that radius of proton is equal to 11 .5...

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