Maple TA 272
Page 1 of 1 • Share
272 #1
Use the equation: DeltaE = (mM)*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 = (mM)*931.5
Plug in the number between the parenthesis and solve. That is the answer.
Answer is negative.
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 = (mM)*931.5
Plug in the number between the parenthesis and solve. That is the answer.
Answer is negative.
DJ Posts : 28
Join date : 20090317
272 #2
Use the same equation (with MeV factor integrated): DeltaE = (mM)*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 = (mM)*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.
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 = (mM)*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.
DJ Posts : 28
Join date : 20090317
Re: Maple TA 272
phyfrk wrote:any luck on question 4
i've tried a few different things but still having no luck
Same here.
DJ Posts : 28
Join date : 20090317
Re: Maple TA 272
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.
DJ Posts : 28
Join date : 20090317
Re:Quest 3
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.
... Guest
274
4. Take the mass given and * 931.5 to get KE in Mev then subtract 0.51MeV to correct for the electron.
Ans. neg
Ans. neg
Yummy Guest
Q#2
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 Guest
Re: Maple TA 272
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?
DJ Posts : 28
Join date : 20090317
please help me with my ques
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).
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).
bo Guest
Q#2
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).
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).
Phent Guest
Q#2
The question asks how much energy is lost per nucleon. Nucleon=proton+neutron.
energy = mass so you must find out how much mass is lost.
Add up all the masses of the proton, neutron, and electron. Subtract that from the actual mass (what is given). That is the mass lost to energy. This is in atomic mass units. You must convert to kg.
1 atomic mass unit = 1.66053886 × 1027 kilograms
e=mc^2 < very important
note that this will be in joules. you need to convert this into Mev/nucleon. So you must convert to Mev and divide by number of nucleons.
1 MeV = 1.6x1013 J
As mass is lost  energy is released so you should know the sign of the answer.
Need a tutor? email el@tantalizingstitches.com
energy = mass so you must find out how much mass is lost.
Add up all the masses of the proton, neutron, and electron. Subtract that from the actual mass (what is given). That is the mass lost to energy. This is in atomic mass units. You must convert to kg.
1 atomic mass unit = 1.66053886 × 1027 kilograms
e=mc^2 < very important
note that this will be in joules. you need to convert this into Mev/nucleon. So you must convert to Mev and divide by number of nucleons.
1 MeV = 1.6x1013 J
As mass is lost  energy is released so you should know the sign of the answer.
Need a tutor? email el@tantalizingstitches.com
El Guest
Question 2 for PHENT
This your question right?
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).
First you do this:
(30*1.007825) + (35*1.008665)= 65.538025
Then:
(65.53802564.924767)*931.5/64.924767 = 8.7986
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).
First you do this:
(30*1.007825) + (35*1.008665)= 65.538025
Then:
(65.53802564.924767)*931.5/64.924767 = 8.7986
Guest786 Guest
Re: Maple TA 272
can you show your work for the 931.5 MeV conversion factor?
Delta E = mass difference * c^2
if the mass difference is 1u (1.66E27) you get
Delta E = 1.66E27*(3E8)^2 = 1.494E10 J
1.494E10J/(1.6E19 J per eV) = 933750000 eV
933750000/ 10E6 = 933.75 MeV
These are the calculations...for some reason I am getting 933.75 instead of the 931.5 quoted in the notes...
TI83 Guest
Page 1 of 1
Permissions in this forum:
You cannot reply to topics in this forum

