Calculating Mass Defect of Americium-244

[Daisy]

Homework Statement

Americium-244 is a rare isotope of Americium. What is the mass defect of Americium-244?

Use the following values for atomic and neutron masses when calculating your answer:

Homework Equations

The Attempt at a Solution

This was what I did:

Mass of protons = 95 x 1.007825 = 95.743375
Isotope = 149 x 1.008665 = 150.291085

Actual Mass Provided = 95.743375 + 150.291085 = 246.03446

Mass Defect = 246.03446 - 244.064279 = 1.970181

I've got that answer and its wrong. I also tried -1.97 and its still wrong. Can somebody please help me?

Do i need to consider electron mass? I computed it to be: Mass Defect with electron mass = 244.064279 - (246.03446 + 0.052155) = 2.022336 which is - 2.022336. Is this correct?

 

[Simon Bridge]

The approach seems correct - computer mediated questions can be finicky about the exact form of the answer.
Your "actual mass provided" seems to have been calculated to only 5dp, but 6dp accuracy provided ...
I get:

> (1.007825*95 + (244-95)*1.008665) - 244.064279
ans =  1.97018099999997

amu for mass deficit.
So it looks like it was fair to round up: you should provide the trailing zero to demonstrate the dp accuracy.
http://physics.bu.edu/~duffy/sc546_notes10/mass_defect.html
... it may well be that you should include the electrons. Did you try? Did you check your notes?

 

[Daisy]

The approach seems correct - computer mediated questions can be finicky about the exact form of the answer.
Your "actual mass provided" seems to have been calculated to only 5dp, but 6dp accuracy provided ...
I get:

> (1.007825*95 + (244-95)*1.008665) - 244.064279
ans =  1.97018099999997

amu for mass deficit.
So it looks like it was fair to round up: you should provide the trailing zero to demonstrate the dp accuracy.
http://physics.bu.edu/~duffy/sc546_notes10/mass_defect.html
... it may well be that you should include the electrons. Did you try? Did you check your notes?

The answers should be left to 3 significant figures. Thats why I can't seem to point out where I went wrong. My lecture notes did not mention anything about it.

 

[Daisy]

Also, do you think the mass defect should be left as a negative answer?

 

[Simon Bridge]

A negative mass deficit would be like a negative deceleration wouldn't it?
If a negative surplus is a deficit then... but check how your course defines it.

Usually you would keep the lowest sig-fig in multiplication ... the lowest would be 2 in the atomic number ... but that's an absolute number so it's really 95.0000 to 6 sig fig. The next lowest is the atomic weight - which is 3 sig fig ... since there may be different isotopes in the sample, one could argue that the sig-fig here is important but IMO that's over-thinking things: nobody uses sig-fig IRL.

Maybe it's just a rule ... it means the computer is testing whether you can guess the format more than it tests your physics.
Fortunately it's not an exam.