Question

An electron is moving with a speed of
3 x 10-7 m/s in a magnetic field of
6x 10-4 T perpendicular to its path. What
will be the radium of the path? What will
be frequency and the energy in keV ?
[Given: mass of electron = 9 x10-31 kg,
1.6 x
charge e = 1.6 10-19 C, 1 eV
10-19 ]​

Answer 1

Given:

An electron is moving with a speed of 3 x 10-7 m/s in a magnetic field of 6x 10-4 T perpendicular to its path.

[Given: mass of electron = 9 x10-31 kg, 1.6 x charge e = 1.6 10-19 C, 1 eV 10-19 ]​

To find:

What will be the radium of the path? What will be frequency and the energy in keV ?

Solution:

From given, we have,

v = 3 x 10-7 m/s

B = 6 x 10-4 T

m = 9 x 10-31 kg

e = 1.6 10-19 C

we use the formulae:

The radius of the path:

evB = mv²/r

∴ r = mv/eB

r = [9 x 10-31 ×  3 x 10-7] / [ 1.6 x 10-19 × 6 x 10-4] = 0.28

∴ The radius of the path is 0.28 m

The frequecy:

F = w/2π = (v/r) × (1/2π)

F = 3 x 10-7 / 0.28 × 1/2π = 1.7 × 10^-7

∴ The frequency is 1.7 × 10^-7 Hz

The energy:

E = 1/2mv²

E  = 1/2 × 9 x 10-31 × 3 x 10-7 = 4.05 × 10^-15

∴ The energy in eV is 4.05 × 10^-15/1.6 10-19 = 25.3 keV

Answer 2

Answer:

From given, we have,

v = 3 x 10-7 m/s

B = 6 x 10-4 T

m = 9 x 10-31 kg

e = 1.6 10-19 C

we use the formulae:

The radius of the path:

evB = mv²/r

∴ r = mv/eB

r = [9 x 10-31 ×  3 x 10-7] / [ 1.6 x 10-19 × 6 x 10-4] = 0.28

∴ The radius of the path is 0.28 m

The frequecy:

F = w/2π = (v/r) × (1/2π)

F = 3 x 10-7 / 0.28 × 1/2π = 1.7 × 10^-7

∴ The frequency is 1.7 × 10^-7 Hz

The energy:

E = 1/2mv²

E  = 1/2 × 9 x 10-31 × 3 x 10-7 = 4.05 × 10^-15

∴ The energy in eV is 4.05 × 10^-15/1.6 10-19 = 25.3 keV

Explanation: