Respuesta :
(a) [tex]3.59\cdot 10^6 m/s[/tex]
The magnetic force acts as centripetal force, so we can write
[tex]qvB= \frac{mv^2}{r}[/tex]
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,
[tex]q=2e=3.2\cdot 10^{-19} C[/tex]
[tex]m=4.00 u = 4\cdot 1.67\cdot 10^{-27} kg=6.68\cdot 10^{-27} kg[/tex]
[tex]r = 6.40 cm = 0.064 m[/tex]
B = 1.17 T
Re-arranging the equation and solving for v, we find its speed:
[tex]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[/tex]
(b) [tex]1.12\cdt 10^{-7} s[/tex]
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
[tex]T= \frac{2\pi r}{v}[/tex]
where
r is the radius of the path
v is the speed
Here we have
[tex]r = 6.40 cm = 0.064 m[/tex]
[tex]v=3.59\cdot 10^6 m/s[/tex]
So the period of revolution is
[tex]T= \frac{2\pi (0.064)}{3.59\cdot 10^6}=1.12\cdt 10^{-7} s[/tex]
(c) [tex]4.30\cdot 10^{-14}J[/tex]
The kinetic energy of a particle is given by
[tex]E_k = \frac{1}{2}mv^2[/tex]
where
m is its mass
v is its speed
For the alpha particle in the problem, we have
[tex]m=4.00 u = 4\cdot 1.67\cdot 10^{-27} kg=6.68\cdot 10^{-27} kg[/tex]
[tex]v=3.59\cdot 10^6 m/s[/tex]
So its kinetic energy is
[tex]E_k = \frac{1}{2}(6.68\cdot 10^{-27}(3.59\cdot 10^6)^2=4.30\cdot 10^{-14}J[/tex]
(d) [tex]1.34\cdot 10^5 V[/tex]
When accelerated through a potential difference, a particle gains a kinetic energy equal to the change in electric potential energy - so we can write:
[tex]q \Delta V = E_k[/tex]
where the term on the left is the change in electric potential energy, with
q is the charge of the particle
[tex]\Delta V[/tex] is the potential difference
Here we have
[tex]q=2e=3.2\cdot 10^{-19} C[/tex] is the charge of the alpha particle
[tex]4.30\cdot 10^{-14}J[/tex] is the kinetic energy
Re-arranging the formula, we find
[tex]\Delta V=\frac{E_k}{q}=\frac{4.30\cdot 10^{-14}}{3.2\cdot 10^{-19}}=1.34\cdot 10^5 V[/tex]
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 :
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]v=\dfrac{qrB}{m}[/tex]
[tex]v=\dfrac{2\times 1.6\times 10^{-19}\times 0.064\times 1.17 }{6.64\times 10^{-27}}[/tex]
[tex]v=3.6\times 10^6\ m/s[/tex]
(b) Period of revolution, [tex]T=\dfrac{2\pi r}{v}[/tex]
[tex]T=\dfrac{2\pi 0.064 }{3.6\times 10^6}[/tex]
[tex]T=1.11\times 10^{-7}\ s[/tex]
(c) Kinetic energy, [tex]E=\dfrac{1}{2}mv^2[/tex]
[tex]E=\dfrac{1}{2}\times 6.64\times 10^{-27}\times (3.6\times 10^6)^2[/tex]
[tex]E=4.3\times 10^{-14}\ J[/tex]
or
[tex]E=268750\ eV[/tex]
(d) Potential energy = Kinetic energy
[tex]2eV=E[/tex]
[tex]2eV=268750\ eV[/tex]
V = 134375 volts
Hence, this is the required solution.
