This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ME311 MATERIALS ENGINEERING HOMEWORK #1 PROBLEM 1: In a collection of carbon atoms, 80% of the atoms contain 6 neutrons and 20% of the atoms contain 8 neutrons. The atomic number for C is 6. Calculate the approximate average atomic mass for this collection of carbon atoms. What is the designation given to atoms of the same element but with a different mass? SOLUTION: The atomic mass of C 12 = number of protons + number of neutrons = 6+6=12 g/mole The atomic mass of C 14 = number of protons + number of neutrons = 6+8=14 g/mole The average atomic mass = 0.8*12 + 0.2*14= 12.4 g/mole Atoms of the same element but with a different mass are designated isotopes. PROBLEM 2: The speed of a bullet (m=50 g) and the speed of an electron (m=9.1x1028 g) are measured to be the same, namely 300m/sec, with an uncertainty of 0.01%. With what fundamental accuracy could we locate the position of each? Hint: Use the Uncertainty Principle of Heisenberg SOLUTION: For the electron p (momentum)=mv = 9.1 x 1031 kg x 300 m/sec = 2.7 x 1028 kg.m/sec and ∆ p= m ∆ v = 0.0001 x 2.7 x1028 kg.m/sec = 2.7 x 1032 kg.m/sec so that ∆ x ≥ h/ ∆ p = (6.6 x 1034 joule.sec)/ (2.7 x 1032 kg.m/sec) = 0.024 m = 2.4 cm For the bullet p (momentum)=mv = 0.05 kg x 300 m/sec = 15 kg.m/secp (momentum)=mv = 0....
View
Full
Document
 Spring '08
 MEYERS
 Atom, Uncertainty Principle, Electrons, average atomic mass, 1s 2s, possible electrons

Click to edit the document details