Maple TA 27-2

Explanations here.

Use the equation: DeltaE = (m-M)*c^2

The difference between the two masses is given already. Now, if you integrate the MeV conversion into the equation as shown on sheet 17, we get:

DeltaE = (m-M)*931.5

Plug in the number between the parenthesis and solve. That is the answer.

Answer is negative.

Use the same equation (with MeV factor integrated): DeltaE = (m-M)*931.5

Now, different numbers of particles are given. Since it is a neutral atom, we assume that the mass of a hydrogen atom can be used for the electrons and protons. However, we will need to add in the mass of the neutrons. So:

(# of protons * 1.007825) + (# of neutrons * 1.008665) = m

Now, plug it into the equation DeltaE = (m-M)*931.5. We just calculated m, and M is given in the question.

This answer, DeltaE, needs to be divided by M because we are trying to find the E per nucleon. That is the answer.

Answer is positive.

The number of protons will remain the same.

Answer is positive.

any luck on question 4
i've tried a few different things but still having no luck

phyfrk wrote:any luck on question 4
i've tried a few different things but still having no luck

Same here.

Actually proton number changes it is Z+1 for the answer. positive.

yo wrote:Actually proton number changes it is Z+1 for the answer. positive.

My posted solution worked for me.

There may be two variations of the question.

Any ideas on number four I am having trouble with it as well.

Maple TA accepted answers from 131 to 135 for me when my number given was 132.. So Z and Z+1 give you the correct answer according to Maple TA.

4. Take the mass given and * 931.5 to get KE in Mev then subtract 0.51MeV to correct for the electron.

Ans. neg

i have trying to follow the equation given above but i keep getting the wrong answer for #2. any help? i get answer like 8.5623

phent wrote:i have trying to follow the equation given above but i keep getting the wrong answer for #2. any help? i get answer like 8.5623

What are your given values?

the method i tried on top doesnt work!!

You "throw together" 30 protons, 35 neutrons and 30 electrons forming a hypothetical atom with an atomic mass of 64.954432 u. What is the binding energy per nucleon in MeV (sheet 13,16) ? Indicate with a positive (negative) sign whether energy is released by (input into) the formation of the atom. (Atomic masses in u: proton 1.007276,neutron 1.008665,electron 0.000549 and neutral H atom 1.007825. Note that the atomic masses (like 64.954432 ) are quoted for the neutral atom. Calculate with the quoted accuracy and enter your answer with the usual 4 significant digits).

DJ.. these are the values.. thanks

You "throw together" 30 protons, 35 neutrons and 30 electrons forming a hypothetical atom with an atomic mass of 64.924767 u. What is the binding energy per nucleon in MeV (sheet 13,16) ? Indicate with a positive (negative) sign whether energy is released by (input into) the formation of the atom. (Atomic masses in u: proton 1.007276,neutron 1.008665,electron 0.000549 and neutral H atom 1.007825. Note that the atomic masses (like 64.924767 ) are quoted for the neutral atom. Calculate with the quoted accuracy and enter your answer with the usual 4 significant digits).