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Fundamentals Of Physics
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Short Answer

a. Show that the mass M of an atom is given approximately by Mapp=Amp, where A is the mass number and is the proton mass. For (b) 1H , (c)31P,(d)120Sn , (e) 197Au, and (f) 239Pu, use Table 42-1 to find the percentage deviation between Mappand M:

role="math" localid="1662047222746" percentage deviation=Mapp-MM×100

(g) Is a value of Mapp accurate enough to be used in a calculation of a nuclear binding energy?

  1. The mass of an atom is given by Mapp=Amp .
  2. The percentage deviation of 1H is -0.05 %.
  3. The percentage deviation of 31P is 0.81%.
  4. The percentage deviation of 120Sn is 0.81%.
  5. The percentage deviation of 197Au is 0.74% .
  6. The percentage deviation of 239Pu is 0.71% .
  7. No, there is no value of accurate enough to be used in a calculation of a nuclear binding energy.
See the step by step solution

Step by Step Solution

Step 1: Write the given data

  1. Given particles: 1H, 31P, 120Sn, 197Au, 239Pu
  2. The formula of percentage deviation, localid="1662047170223" =Mapp-MM×100

Step 2: Determine the concept of mass  

The nucleus of an atom is made up of protons and neutrons that together are called nucleons. Thus, the mass of an atom can be determined on that basis as the mass number is defined as the sum of the number of nucleons. Now, using the given percentage mass deviation formula, we can get the required values of the deviations of the element.

Formulae:

The mass number of an element as follows:

A = n + p …… (i)

The percentage deviation from the mass is as follows:

percetage deviation=Mapp-MM×100 …… (ii)

Step 3: a) Determine the mass of an atom

The mass number A is the number of nucleons in an atomic nucleus. Since mpmn , the mass of the nucleus is approximately is given using equation (i) as follows:

Mapp=pmp+nmn =p+nmp =Amp

Also, the mass of the electrons is negligible since it is much less than that of the nucleus.

Hence, the mass of an atom is given by Mapp=Amp , where is the mass of the proton.

Step 4: b) Calculate the percentage deviation H1

From the calculations of part (a), we get the apparent mass from equation for as follows:

Mapp=11.007276u =1.007276u mp=1.007276u

The actual mass of 1His M = 1.007825u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

percentage deviation=1.007276u-1.007825u1.007825u×100 =-0.054% -0.05%

Hence, the value of the deviation is -0.05% .

Step 5: c) Calculate the percentage deviation P31

From the calculations of part (a), determine apparent mass from equation (a) as follows:

Mapp=11.007276u =31.225556u mp=1.007276u

The actual mass of 31Pis M=30.973762u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

percentage deviation=31.225556-30.973762u30.973762u×100 =0.81%

Hence, the value of the deviation is 0.81% .

Step 6: d) Calculate the percentage deviation Sn120

From the calculations of part (a), we get the apparent mass from equation (a) as follows:

Mapp=(120)1.007276u mp=1.007276u =120.87312u

The actual mass of 120Snis M=119.902197u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

percentage deviation=120.87312u-119.902197u119.902197u×100 =0.81%

Hence, the value of the deviation is 0.81% .

Step 7: e) Calculate the percentage deviation Au197

From the calculations of part (a), we get the apparent mass from equation as follows:

Mapp=(197)1.007276u mp=1.007276u =198.433372u

The actual mass of 197Au is M = 196.96652u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

percentage deviation=198.433372u-196.966552u196.966552u×100 =0.74%

Hence, the value of the deviation is 0.74% .

Step 8: f) Calculate the percentage deviation Pu239

From the calculations of part (a), we get the apparent mass from equation as follows:

Mapp=(239)1.007276u mp=1.007276u =240.738964u

The actual mass of 239Puis M = 230.052157u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

percentage deviation=240.738964u-239.052157u239.052157u×100 =0.71%

Hence, the value of the deviation is 0.71 %.

Step 9: g) Calculate the value of the apparent mass that can be used for nuclear binding energy calculations

No. In a typical nucleus the binding energy per nucleon is several MeV, which is a bit less than 1% of the nucleon mass times c2. This is comparable with the percent error calculated in parts (b) – (f), so we need to use a more accurate method to calculate the nuclear mass.

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