# Calculate the radius of curvature of the path of a particle.

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1. Calculate the radius of curvature of the path of a beta particle of mass 9.1 x 10^-31 kg and speed 2.0 x 10^7 m/sec perpendicular to a magnetic field of 0.50 T.

a. 0.23 mm

b. .46 mm

c. 0.92 mm

d. 0.051 mm

2. Calculate the radius of curvature of the path of an alpha particle of mass 6.6 x 10^-27 kg moving with a kinetic energy of 8.7 x 10^-14 J in a magnetic field of 1.00 T.

a. 0.22 m

b. 0.11 m

c. .055 m

d. .44 m

3. A sample of polonium has a half-life of 3.0 min and an intial activity of 10 x 10^6 Bq. What is its activity after 30 min have elapsed?

a. 9800 Bq

b. 4600 Bq

c. 19,600 Bq

d. 22,000 Bq

4. The ratio of radioactive to nonradioactive carbon in a sample of wood is found to be one-third that found in a recently cut piece of wood. Approximately how long ago was the first sample cut? The half-life of carbon-14 is 5700 years.

a. 9000 y

b. 4500 y

c. 18,000 y

d. 2250 y

5. What is the distance of closest approach of an alpha particle of speed v= 1.70 x10^7 m/s to a gold nucleus of Z = 79? How close would the alpha get to a copper nucleus of Z = 29? Assume that the gold and copper nuclei remain at rest.

a. a) 3.8 x 10^(-14) m b) 1.4 x 10^(-14) m

b. a) 1.9 x 10^(-14) m b) 0.7 x 10^(-14) m

c. a) 7.6 x 10^(-14) m b) 2.8 x 10^(-14) m

d. a) 2.8 x 10^(-14) m b) 2.4 x 10^(-14) m

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1.)

m = 9.1 x 10^-31 kg

speed v = 2.0 x 10^7 m/sec

magnetic field B = 0.50 T.

charge q = 1.6)16(-19) C

Lorentz force F = q*v*B = m*v^2/r

=> r = m*v/(q*B)

=> r = 9.1 x 10^(-31)*2.0 x 10^7/(1.6*10^(-19)*0.5)

=> r = 0.0002275 m = 0.2275 mm = 0.23 mm --Answer (a)

2.)

radius of curvature r = ?

mass m = 6.6 x 10^-27 kg

kinetic energy E = 0.5*m*v^2 = 8.7 x 10^-14 J

magnetic field B = 1.00 T

charge q = 4*1.6*10^(-19) C = 6.4*10^(-19) ...

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- BEng, Allahabad University, India
- MSc , Pune University, India
- PhD (IP), Pune University, India

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