How Were the Atomic Weights Calculated in Question 50?

[ZedCar]

Homework Statement

In question 50 in this link:

http://www.tarleton.edu/Faculty/alow/prob21.htm

I was wondering how the numbers 4.00260 amu + 1.00867 amu – 2.01410 amu – 3.01605 amu been obtained?

Thank you.

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Hi ZedCar!

I was wondering how the numbers 4.00260 amu + 1.00867 amu – 2.01410 amu – 3.01605 amu been obtained?

that's atomic mass units (with carbon-12 defined as 12.000000) …

they probably got it from a poster on the lab wall, much like this one … http://en.wikipedia.org/wiki/Atomic_weight#Periodic_table_by_atomic_weight

ultimately, i suppose they did it by weighing them ​

 

[ZedCar]

Thank you.

The first part of question 50 states 2H + 3H → 4He + 1n

So are they multiplying these numbers by a certain constant to arrive at the numbers of 4.00260 amu + 1.00867 amu – 2.01410 amu – 3.01605 amu

If so, what is this constant?

As the numbers do look just slightly more than the numbers which I have emboldened.

 

[tiny-tim]

they're not a constant times anything, atomic weight is not a linear function of the number of nucleons

fuse two protons (¹H) together, that's two lots of 1.00867 = 2.01734, you get a ²H (= 2.01410) and some energy left over, which is the energy of fusion that they're trying to make use of in fusion reactors

 

[Borek]

you get a ²H

You may want to change it to something closer to reality

 

[tiny-tim]

oops!