An electron is moving with a speed of 3 ×10^7 m/s
in a magnetic field of 6 x 10^-4T perpendicular to
its path. What will be the radius of the path? What
will be frequency and energy in keV?
(Given: m =9.1 x 10^31 kg, e=1.6% 10^-19
Answers
Answered by
1
Answer:
Here, m=9×10−31kg,
q=1.6×10−19C,v=3×107ms−1,
b=6×10−4T
r=qBmv=(1.6×10−19)(6×10−4)(9×10−31)×(3×107)=0.28m
v=2πrv=2πmBq=2×(22/7)×9×10−31(6×10−4)×(1.6×10−19)
=1.7×107Hz
Ek=21mv2=
21×(9×10−31)×(3×107)
Answered by
1
Answer:
Here, m=9×10
−31
kg,
q=1.6×10
−19
C,v=3×10
7
ms
−1
,
b=6×10
−4
T
r=
qB
mv
=
(1.6×10
−19
)(6×10
−4
)
(9×10
−31
)×(3×10
7
)
=0.28m
v=
2πr
v
=
2πm
Bq
=
2×(22/7)×9×10
−31
(6×10
−4
)×(1.6×10
−19
)
=1.7×10
7
Hz
Ek=
2
1
mv
2
=
2
1
×(9×10
−31
)×(3×10
7
)
2
J
=40.5×10
−17
J=
1.6×10
−16
40.5×10
−17
keV
=2.53keV
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