Respuesta :
Explanation:
Given that,
The kinetic energy of the electron, [tex]K=1.67\ keV=1.67\times 10^3\ eV[/tex]
Radius of the orbit, r = 37.4 cm = 0.374 m
To find,
(a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.
Solve,
We know that, [tex]1\ eV=1.6\times 10^{-19}\ J[/tex]
[tex]K=1.67\times 10^3\times 1.6\times 10^{-19}\ J=2.67\times 10^{-16}\ J[/tex]
(a) The formula of the kinetic energy is given by :
[tex]K=\dfrac{1}{2}mv^2[/tex]
v is the speed if electron
m is the mass of electron
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 2.67\times 10^{-16}}{9.1\times 10^{-31}}}[/tex]
[tex]v=2.42\times 10^7\ m/s[/tex]
(b) Let B is the magnitude of magnetic field. On the circular path the magnetic field is given by :
[tex]B=\dfrac{mv}{qr}[/tex]
[tex]B=\dfrac{9.1\times 10^{-31}\times 2.42\times 10^7}{1.6\times 10^{-19}\times 0.374}[/tex]
[tex]B=3.68\times 10^{-4}\ T[/tex]
(c) Let T is the time period of the motion. It is given by :
[tex]T=\dfrac{2\pi r}{v}[/tex]
Circling frequency is given by :
[tex]f=\dfrac{1}{T}=\dfrac{v}{2\pi r}[/tex]
[tex]f=\dfrac{2.42\times 10^7}{2\pi \times 0.374 }[/tex]
[tex]f=1.02\times 10^7\ Hz[/tex]
(d) The period of motion,
[tex]T=\dfrac{1}{f}[/tex]
[tex]T=\dfrac{1}{1.02\times 10^7}[/tex]
[tex]T=9.80\times 10^{-8}\ s[/tex]
