Question

1. A gold atom has a mass of 3.287 • 10-25 kg and is at rest. It emits an alpha particle that has a mass of 6.646 • 10-27 kg. In this process, the gold atom is transformed into an iridium isotope. The alpha particle has a speed of 3.40 • 106 m/s.

A. What is the absolute value of the momentum of the alpha particle?
B. What is the absolute value of the momentum of the iridium isotope after the emission of the alpha particle?
C. What is the speed of the iridium isotope after the emission process?

Hint: For the purposes of this problem, assume that the mass of the iridium isotope is simply the difference of the mass of the gold atom and that of the alpha particle.

Answer 1

m₁=6.646•10⁻²⁷ kg, v₁=3.4•10⁶ m/s,
m₂=3.287•10⁻²⁵-6.646•10⁻²⁷=3.22054•10⁻²⁵ kg,
v₂=?

p₁=m₁v₁=6.646•10⁻²⁷•3.4•10⁶=2.25964•10⁻²⁰ kg•m/s
The law of conservation of linear momentum
0 =m₁v₁-m₂v₂
v₂=m₁v₁/m₂=
=6.646•10⁻²⁷•3.4•10⁶/3.22054•10⁻²⁵=7.02•10⁴ m/s
p₂=m₂v₂=…

Answer 2

Thank you so much!
I got 22.61x10^-21 is that right?

Answer 3

My only question is how to find the speed of the particles