A proton,an electron, and a Helium nucleus, have the same energy. they are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let rp , re and rHe be their respective radii, then,a)re rp = rHeb)re rp rHec)re

Question

A proton,an electron, and a Helium nucleus, have the same energy. they are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let rp , re and rHe be their respective radii, then,a)re > rp = rHeb)re > rp > rHec)re < rp < rHed)re < rp = rHeCorrect answer is option 'D'. Can you explain this answer?

Answer

Note :

- Here P denotes momentum of a particle

- Mass of helium = 4 x mass of proton

- Charge on helium = 2 x charge on proton

- Charge on electron = charge of proton but mass of electron < mass of proton

Answer

Answer

The radius of a charged particle's circular orbit in a magnetic field is given by the equation:

r = (mv)/(|q|B)

Where r is the radius of the orbit, m is the mass of the particle, v is the velocity of the particle, q is the charge of the particle, and B is the magnetic field strength.

Since the proton, electron, and helium nucleus all have the same energy, we can assume that they have the same velocity. Therefore, the radii of their orbits will depend on their masses and charges.

The mass of a proton (mp) is approximately 1.67 x 10^-27 kg, the charge of a proton (qp) is +1.6 x 10^-19 C, and the charge-to-mass ratio (qp/mp) is approximately 9.58 x 10^7 C/kg.

The mass of an electron (me) is approximately 9.11 x 10^-31 kg, the charge of an electron (qe) is -1.6 x 10^-19 C, and the charge-to-mass ratio (qe/me) is approximately -1.76 x 10^11 C/kg.

The mass of a helium nucleus (mHe) is approximately 6.64 x 10^-27 kg, the charge of a helium nucleus (qHe) is +3.2 x 10^-19 C, and the charge-to-mass ratio (qHe/mHe) is approximately 4.82 x 10^7 C/kg.

Since the radius of the orbit is inversely proportional to the charge-to-mass ratio, we can compare the radii of the orbits as follows:

re = (me * v)/(|qe| * B)
rp = (mp * v)/(|qp| * B)
rHe = (mHe * v)/(|qHe| * B)

Since the velocity (v) is the same for all three particles, we can cancel it out when comparing the radii:

re/rp = (mp * |qe|)/(me * |qp|)
rHe/rp = (mp * |qHe|)/(mHe * |qp|)

Substituting the values for the masses and charges:

re/rp = (1.67 x 10^-27 kg * 1.6 x 10^-19 C)/(9.11 x 10^-31 kg * 1.6 x 10^-19 C)
re/rp ≈ 1.84 x 10^7

rHe/rp = (1.67 x 10^-27 kg * 3.2 x 10^-19 C)/(6.64 x 10^-27 kg * 1.6 x 10^-19 C)
rHe/rp ≈ 3.20

Therefore, the ratio of the radius of the electron's orbit to the radius of the proton's orbit is approximately 1.84 x 10^7, and the ratio of the radius of the helium nucleus's orbit to the radius of the proton's orbit is approximately 3.20.

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