Respuesta :
Answer:
(a) [tex]v=22.177\times 10^{12}\ m.s^{-1}[/tex]
(b) [tex]B=420.855\ T[/tex]
(c) [tex]f=11.7647\times 10^{14}\ Hz[/tex]
(d) [tex]T=8.5\times 10^{-14}s[/tex]
Explanation:
Given:
Kinetic Energy of an electron, [tex]KE=1.4\ keV=1400\ eV=1400\times 1.6\times 10^{-19}\ J[/tex]
radius of the orbit, [tex]r=0.3\ m[/tex]
we have:
mass of an electron, [tex]m=9.109\times 10^{-31}\ kg[/tex]
charge on an electron, [tex]q=1.6\times 10^{-19}\ C[/tex]
(a)
we know:
[tex]KE=\frac{1}{2} m.v^2[/tex]
[tex]1400\times 1.6\times 10^{-19}=0.5\times (9.109\times 10^{-31})\times v^2[/tex]
[tex]v=22.177\times 10^{12}\ m.s^{-1}[/tex]
(b)
We also have the relation after the comparison of forces(centripetal and magnetic) on a moving charge in a magnetic field as:
[tex]m.v =q.B.r[/tex] ...........................(1)
where:
B = magnetic field normal to the plane of circulating charge
putting respective values in eq. (1)
[tex](9.109\times 10^{-31})\times (22.177\times 10^{12})=(1.6\times 10^{-19})\times 0.3\times B[/tex]
[tex]B=420.855\ T[/tex]
(d)
angular speed:
[tex]\omega=\frac{v}{r}[/tex]
[tex]\omega=\frac{22.177\times 10^{12}}{0.3}[/tex]
[tex]\omega=73.92\times 10^{12}\ rad.s^{-1}[/tex]
∴Time taken for 1 radian:
[tex]t=\frac{1}{\omega}[/tex]
[tex]t=\frac{1}{73.92\times 10^{12}}\ s.rad^{-1}[/tex]
Now time take for 1 circulation i.e. 2π radians(Time period):
[tex]T=2\pi\times t[/tex]
[tex]T=2\pi\times \frac{1}{73.92\times 10^{12}}[/tex]
[tex]T=8.5\times 10^{-14}s[/tex]
(c)
we know frequency :
[tex]f=\frac{1}{T}[/tex]
[tex]f=\frac{1}{8.5\times 10^{-14}}[/tex]
[tex]f=11.7647\times 10^{14}\ Hz[/tex]
