1 answer

An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of...

Question:

An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.13 cm in a uniform magnetic field with B = 1.62 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.


Answers

(a)

qvB=m\frac{v^{2}}{r}

v= \frac{qBr}{m}=\frac{(+2(1.6\times 10^{-19}))(1.62T)(0.0513m)}{4.00(1.67\times 10^{-27})}=3.98\times 10^{6}m/s

(b)

period

T = \frac{2\pi r}{v}=\frac{2\pi (0.0513)}{3.98\times 10^{6}}=8.09\times 10^{-8}s

(c)

Kinetic energy

=\frac{1}{2}mv^{2}=\frac{1}{2}(4.00)(1.67\times 10^{-27})(3.98\times 10^{6})=1.33\times 10^{-20}J=0.083eV

(d)

qV = K.E. = 1.33\times 10^{-20}J

V = \frac{1.33\times 10^{-20}}{2(1.6\times 10^{-19})}=0.042V

.

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