Physics, 12.08.2019 18:10 omojay1257

An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 6.40 cm in a uniform magnetic field with b = 1.17 t. calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

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Answer from: Lovescats9191

(a) $3.59\cdot 10^6 m/s$

The magnetic force acts as centripetal force, so we can write

$qvB= \frac{mv^2}{r}$

where

q is the charge of the particle

v is its speed

B is the magnetic field strength

m is the mass

r is the radius of the circular path

For the alpha particle in the problem,

$q=2e=3.2\cdot 10^{-19} C$

$m=4.00 u = 4\cdot 1.67\cdot 10^{-27} kg=6.68\cdot 10^{-27} kg$

r = 6.40 cm = 0.064 m

B = 1.17 T

Re-arranging the equation and solving for v, we find its speed:

$v=\frac{qBr}{m}=\frac{(3.2\cdot 10^{-19})(1.17)(0.064 m)}{6.68\cdot 10^{-27}}=3.59\cdot 10^6 m/s$

(b) $1.12\cdt 10^{-7} s$

The period of revolution is given by the ratio between the distance travelled in one circle (so, the circumference of the path) and the speed of the particle, so

$T= \frac{2\pi r}{v}$

where

r is the radius of the path

v is the speed

Here we have

r = 6.40 cm = 0.064 m

$v=3.59\cdot 10^6 m/s$

So the period of revolution is

$T= \frac{2\pi (0.064)}{3.59\cdot 10^6}=1.12\cdt 10^{-7} s$

The kinetic energy of a particle is given by

$E_k = \frac{1}{2}mv^2$

where

m is its mass

v is its speed

For the alpha particle in the problem, we have

$m=4.00 u = 4\cdot 1.67\cdot 10^{-27} kg=6.68\cdot 10^{-27} kg$

$v=3.59\cdot 10^6 m/s$

So its kinetic energy is

$E_k = \frac{1}{2}(6.68\cdot 10^{-27}(3.59\cdot 10^6)^2=4.30\cdot 10^{-14}J$

(d) $1.34\cdot 10^5 V$

When accelerated through a potential difference, a particle gains a kinetic energy equal to the change in electric potential energy - so we can write:

$q \Delta V = E_k$

where the term on the left is the change in electric potential energy, with

q is the charge of the particle

$\Delta V$ is the potential difference

Here we have

$q=2e=3.2\cdot 10^{-19} C$ is the charge of the alpha particle

$4.30\cdot 10^{-14}J$ is the kinetic energy

Re-arranging the formula, we find

$\Delta V=\frac{E_k}{q}=\frac{4.30\cdot 10^{-14}}{3.2\cdot 10^{-19}}=1.34\cdot 10^5 V$

Answer from: Emptypockets451

Explanation:

It is given that,

Charge on alpha, q = 2e

Radius of circular path, r = 6.4 cm = 0.064 m

Mass of alpha particle, m = 4 u

Uniform magnetic field, B = 1.17 T

(a) Magnetic force is balanced by the centripetal force as :

$qvB=\dfrac{mv^2}{r}$

$v=\dfrac{qrB}{m}$

$v=\dfrac{2\times 1.6\times 10^{-19}\times 0.064\times 1.17 }{6.64\times 10^{-27}}$

$v=3.6\times 10^6\ m/s$

(b) Period of revolution, $T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi 0.064 }{3.6\times 10^6}$

$T=1.11\times 10^{-7}\ s$

$E=\dfrac{1}{2}\times 6.64\times 10^{-27}\times (3.6\times 10^6)^2$

$E=4.3\times 10^{-14}\ J$

$E=268750\ eV$

(d) Potential energy = Kinetic energy

2eV=E

$2eV=268750\ eV$

V = 134375 volts

Hence, this is the required solution.

Answer from: Quest

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Answer from: Quest

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