This questions involves youngs modulus

Question

A mole of gold atoms are arranged in a cube. A 1 kg mass is placed on top of the cube. The interatomic spacing for gold is 10^−10 m andYoung’s modulus of gold is 1011 N/m2. Given there are roughly 10^24 atoms in one mole and the acceleration due to gravity is 10 m/s2 how much does the cube compress?

 

Answer choices 

 

1. 10−10 m
2. 10−4 m
3. 10−16 m
4. 10−14 m
5. 10−12 m
6. 10−8 m
7. 10−6 m

 

Answer 1

An even or uniform distribution of 10²⁴ Gold atoms in a perfect cubic space, with 10^-10 meters of horizontal or vertical distance between the Gold atoms, would give a cube with a side of (10^-10 meters per space times

(10⁸ − 1) spaces) equal to 10^-2 meters. Each cube face would be (0.01)² square meters in area.

1 kilogram placed on top of the cube with the acceleration due to Earth's gravity would exert a downward force

on the Gold cube equal to 1 kilogram times 9.8 meters per second every second or 9.8 (kilogram-dot-meters-per-second-squared) or 9.8 Newtons. The average Young's Modulus for Gold can be taken as 78.5×10⁹ Pascals (Newtons per square meter).

Then write 78.5×10⁹ = (9.8/(0.01)²) ÷ (ΔL/(0.01)) and isolate ΔL as [(9.8/(0.01)²) ÷ 78.5×10⁹](0.01) or 1.248407643×10^-8 meters equivalent to 10^-8 meters for length of compression.