Asked by sam
1. When material decays, it emits a particle. An experiment similar to the J.J. Thomson experiment is performed to determine the charge-to-mass ratio of this particle. It is found that the particle moves undeflected through mutually perpendicular magnetic and electric fields of 2.00 × 10^ -3 T and 1.08 × 10^4 N/C, respectively. When the electric field is turned off, the particle is found to deflect to a radius of 1.53 × 10^ -2m. determine the type of particle emitted. Show all the steps needed to make this determination.
2.Using your periodic table, write the decay equation for the decay of the previous question, identifying the product isotope.
3.How would a gamma ray have been affected by passing through a magnetic field as mentioned in the first question?
2.Using your periodic table, write the decay equation for the decay of the previous question, identifying the product isotope.
3.How would a gamma ray have been affected by passing through a magnetic field as mentioned in the first question?
Answers
Answered by
Elena
1.
F(el) =F(magn)
qE=qvB
v=E/B=1.08•10⁴/2•10⁻³ =5.4•10⁶ m/s
ma=mv²/R= qvB
q/m=v/RB =5.4•10⁶/1.53•10⁻²•2•10⁻³ =
=1.76•10¹¹ C/kg +>
This is electron (e/m= 1.6•10⁻¹⁹/9.1•10⁻³¹=1.76•10¹¹ C/kg)
2.
X (A,z) -> e(0, -1) +Y(A, z-1)
3.
Gamma rays are photons(Electromagnetic radiation). Photons are particles with no mass,
no charge and no magnetic moment. Hence they cannot be affected by either an electric or a magnetic field.
F(el) =F(magn)
qE=qvB
v=E/B=1.08•10⁴/2•10⁻³ =5.4•10⁶ m/s
ma=mv²/R= qvB
q/m=v/RB =5.4•10⁶/1.53•10⁻²•2•10⁻³ =
=1.76•10¹¹ C/kg +>
This is electron (e/m= 1.6•10⁻¹⁹/9.1•10⁻³¹=1.76•10¹¹ C/kg)
2.
X (A,z) -> e(0, -1) +Y(A, z-1)
3.
Gamma rays are photons(Electromagnetic radiation). Photons are particles with no mass,
no charge and no magnetic moment. Hence they cannot be affected by either an electric or a magnetic field.
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