Difference between revisions of "Specific Charge"
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There are 22 [[proton]]s in the [[Atomic Nucleus|nucleus]] of [[Titanium]]. | There are 22 [[proton]]s in the [[Atomic Nucleus|nucleus]] of [[Titanium]]. | ||
− | <math>Q = 22\ | + | <math>Q = 22\times1.60\times10^{-19}</math>C |
<math>m = (z + n)\times1.67\times10^{-27}</math>kg | <math>m = (z + n)\times1.67\times10^{-27}</math>kg | ||
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<math>S.C. = \frac{Q}{m}</math> | <math>S.C. = \frac{Q}{m}</math> | ||
− | <math>4.58\times10^{7} = \frac{22\ | + | <math>4.58\times10^{7} = \frac{22\times1.60\times10^{-19}}{(22 + n)\times1.67\times10^{-27}}</math> |
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].''' | ||
Line 207: | Line 207: | ||
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] and [[Evaluate (Maths)|evaluate]] the equation, then [[Solve (Maths)|solve]].''' | | style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] and [[Evaluate (Maths)|evaluate]] the equation, then [[Solve (Maths)|solve]].''' | ||
− | <math>(22 + n) = \frac{22\times\1. | + | <math>(22 + n) = \frac{22\times\1.60times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}}</math> |
<math>n = \frac{22\times\1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}} - 22</math> | <math>n = \frac{22\times\1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}} - 22</math> |
Revision as of 19:37, 28 July 2019
Contents
Key Stage 5
Meaning
Specific charge is the ratio of the charge of a particle to its mass shown in the equation \(S.C. = \frac{Q}{m}\) where 'Q' is the charge of the particle and 'm' is the mass of the particle.
About Specific Charge
- The SI Units for specific charge are the Coulomb per kilogram (Ckg-1).
- The specific charge of a particle is a useful quantity when using a mass spectrometer or cloud chamber as it determines the rate of curvature of a particle with a given its velocity through a known magnetic flux density.
Equation
\(S.C. = \frac{Q}{m}\)
Where
\(S.C. =\) The specific charge of a particle.
\(Q =\) The charge of the particle.
\(m =\) The mass of the particle.
Example Calculations
Calculating Specific Charge
Calculate the specific charge of a proton. | Calculate the specific charge of a Magnesium-24 2+ ion. |
1. State the known quantities in SI Units.
\(Q_{proton} = 1\times1.60\times10^{-19}\)C \(m_{proton} = 1\times1.67\times10^{-27}\)kg |
1. State the known quantities in SI Units.
The charge on an Mg 2+ ion is 2 x elementary charge. \(Q_{Mg2+} = 2\times1.60\times10^{-19}\)C The mass of Magnesium-24 2+ ion is the mass of 24 nucleons + 10 electrons. \(m_{Mg24} = (24\times1.67\times10^{-27} + 10\times9.11\times10^{-31})\)kg |
2. Substitute the numbers and evaluate.
\(S.C. = \frac{Q}{m}\) \(S.C. = \frac{1\times1.60\times10^{-19}}{1\times1.67\times10^{-27}}\) |
2. Substitute the numbers and evaluate.
\(S.C. = \frac{Q}{m}\) \(S.C. = \frac{2\times1.60\times10^{-19}}{24\times1.67\times10^{-27} + 10\times9.11\times10^{-31}}\) |
3. solve the equation.
\(S.C. = \frac{Q}{m}\) \(S.C. = \frac{1.60\times10^{-19}}{1.67\times10^{-27}}\) \(S.C. = 9.58\times10^{7}Ckg^{-1}\) Correct to 3 significant figures. |
3. solve the equation.
\(S.C. = \frac{Q}{m}\) \(S.C. = \frac{3.20\times10^{-19}}{4.01\times10^{-26}}\) \(S.C. = 7.98\times10^{6}Ckg^{-1}\) Correct to 3 significant figures. |
Calculating Charge
Calculate the charge of a Cobalt ion with a specific charge of 4.87x106Ckg-1 and a mass of 9.86x10-26kg. | Calculate the charge of a nucleus with a specific charge of 4.51x107Ckg-1 and a mass of 8.52x10-26kg and identify the element. |
1. State the known quantities in SI Units.
\(S.C._{Co} = 4.87\times10^{6}\)Ckg-1 \(m_{Co} = 9.86\times10^{-26}\)kg |
1. State the known quantities in SI Units.
\(S.C. = 4.51\times10^{7}\)Ckg-1 \(m = 8.52\times10^{-26}\)kg |
2. Substitute the numbers and evaluate.
\(S.C. = \frac{Q}{m}\) \(4.87\times10^{6} = \frac{Q}{9.86\times10^{-26}}\) |
2. Substitute the numbers and evaluate.
\(S.C. = \frac{Q}{m}\) \(4.51\times10^{7} = \frac{Q}{8.52\times10^{-26}}\) |
3. Rearrange the equation and solve.
\(Q = 4.87\times10^{6} \times 9.86\times10^{-26}\) \(Q = 4.80\times10^{-19}\)C |
3. Rearrange the equation and solve.
\(Q = 4.51\times10^{7} \times 8.52\times10^{-26}\) \(Q = 3.84\times10^{-18}\)C To identify the nucleus divide the charge by the elementary charge to find the number of protons. \(Z = \frac{3.84\times10^{-18}}{1.60\times10^{-19}}\) \(Z = 24.0\) There are 24 protons in the nucleus therefore the element is Chromium. |
Calculating Mass
Calculate the mass of an ion with an charge of -3.20x10-19C and a specific charge of -5.80x106Ckg-1. | Calculate the relative atomic mass of a nucleus with an charge of 9.61x10-19C and a specific charge of 4.11x107Ckg-1. |
1. State the known quantities in SI Units.
\(S.C. = -5.80\times10^{6}\)Ckg-1 \(Q = -3.20\times10^{-19}\)C |
1. State the known quantities in SI Units.
\(S.C. = 4.11\times10^{7}\)Ckg-1 \(Q = 9.61\times10^{-19}\)C |
2. Substitute the numbers and evaluate.
\(S.C. = \frac{Q}{m}\) \(-5.80\times10^{6} = \frac{-3.20\times10^{-19}}{m}\) |
2. Substitute the numbers and evaluate.
\(S.C. = \frac{Q}{m}\) \(4.11\times10^{7} = \frac{9.61\times10^{-19}}{m}\) |
3. Rearrange the equation and solve.
\(m=\frac{-3.20\times10^{-19}}{-5.80\times10^{6}}\) \(m=5.52\times10^{-26}kg\) |
3. Rearrange the equation and solve.
\(m=\frac{9.61\times10^{-19}}{4.11\times10^{7}}\) \(m=2.34\times10^{-26}kg\) To find the atomic mass divide this mass by the mass of a single nucleon. \(A=\frac{2.34\times10^{-26}}{1.67\times10^{-27}}\) \(A=14.0\) The relative atomic mass is 14. |
Calculating the number of Neutrons
Calculate the number of neutrons in an isotope of Titanium with a nucleus of specific charge 4.58x107Ckg-1. | Calculate the number of neutrons in an isotope of Neon with the specific charge of ....... for a Neon 2- ion. |
1. State the known quantities in SI Units.
\(S.C. = 4.58\times10^{7}\)Ckg-1 There are 22 protons in the nucleus of Titanium. \(Q = 22\times1.60\times10^{-19}\)C \(m = (z + n)\times1.67\times10^{-27}\)kg Where z is the number of protons and n is the number of neutrons and (z+n) is the relative atomic mass. Therefore \(m = (22 + n)\times1.67\times10^{-27}\)kg |
1. State the known quantities in SI Units. |
2. Substitute the numbers and evaluate.
\(S.C. = \frac{Q}{m}\) \(4.58\times10^{7} = \frac{22\times1.60\times10^{-19}}{(22 + n)\times1.67\times10^{-27}}\) |
2. Substitute the numbers and evaluate. |
3. Rearrange and evaluate the equation, then solve.
\((22 + n) = \frac{22\times\1.60times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}}\) \(n = \frac{22\times\1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}} - 22\) \(n = 46.0 - 22\) \(n = 24.0\) |
3. Rearrange the equation and solve. |
Calculating Atomic Number
Calculate the number of protons in a atomic nucleus of an isotope with a mass of ..... and a specific charge of ....... | Calculate the number of protons in a atomic nucleus of an isotope with a mass of ..... and the specific charge of ....... for its 5+ ion. |
1. State the known quantities in SI Units. | 1. State the known quantities in SI Units. |
2. Substitute the numbers and evaluate. | 2. Substitute the numbers and evaluate. |
3. Rearrange the equation and solve. | 3. Rearrange the equation and solve. |