# Maple TA 27-2

## 27-2 #1

Use the equation:

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:

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

Answer is negative.

**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.

**DJ**- Posts : 28

Join date : 2009-03-17

## 27-2 #2

Use the same equation (with MeV factor integrated):

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:

Now, plug it into the equation

This answer, DeltaE, needs to be divided by

Answer is positive.

**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.

**DJ**- Posts : 28

Join date : 2009-03-17

## Re: Maple TA 27-2

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 : 2009-03-17

## Re: Maple TA 27-2

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 : 2009-03-17

## 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

## 27-4

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 27-2

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 : 2009-03-17

## 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 × 10-27 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 = 1.6x10-13 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 × 10-27 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 = 1.6x10-13 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:

(65.538025-64.924767)*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:

(65.538025-64.924767)*931.5/64.924767 = 8.7986

**Guest786**- Guest

## Re: Maple TA 27-2

can you show your work for the 931.5 MeV conversion factor?

Delta E = mass difference * c^2

if the mass difference is 1u (1.66E-27) you get

Delta E = 1.66E-27*(3E8)^2 = 1.494E-10 J

1.494E-10J/(1.6E-19 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:**

**cannot**reply to topics in this forum