SolutionHW1-new

SolutionHW1-new - ME311 MATERIALS ENGINEERING HOMEWORK#1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the 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=9.1x10-28 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 10-31 kg x 300 m/sec = 2.7 x 10-28 kg.m/sec and ∆ p= m ∆ v = 0.0001 x 2.7 x10-28 kg.m/sec = 2.7 x 10-32 kg.m/sec so that ∆ x ≥ h/ ∆ p = (6.6 x 10-34 joule.sec)/ (2.7 x 10-32 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

{[ snackBarMessage ]}

Page1 / 5

SolutionHW1-new - ME311 MATERIALS ENGINEERING HOMEWORK#1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online