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(Created page with "==Key Stage 5== ===Meaning=== '''Specific charge''' is the ratio of the charge of a particle to its mass shown in the equation <math>S.C. = \...")
 
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===About Specific Charge===
 
===About Specific Charge===
 
: The [[SI Unit]]s for '''specific charge''' are the [[Coulomb]] per [[kilogram]] (Ckg<sup>-1</sup>).
 
: The [[SI Unit]]s for '''specific charge''' are the [[Coulomb]] per [[kilogram]] (Ckg<sup>-1</sup>).
−
: The '''specific charge''' of a [[particle]] is a useful quantity when using a [[Mass Spectrometer|mass spectrometer]] or [[Cloud Chamber|cloud chamber]] as it determines the rate of curvature of a [[particle]] with a given [[velocity]] through a known [[Magnetic Field Strength|magnetic field strength]].
+
: The '''specific charge''' of a [[particle]] is a useful quantity when using a [[Mass Spectrometer|mass spectrometer]] or [[Cloud Chamber|cloud chamber]] as it determines the rate of curvature of a [[particle]] with a given its [[velocity]] through a known [[Magnetic Flux Density|magnetic flux density]].
  
 
===Equation===
 
===Equation===
Line 20: Line 20:
 
===Example Calculations===
 
===Example Calculations===
 
====Calculating Specific Charge====
 
====Calculating Specific Charge====
 +
{| class="wikitable"
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the '''specific charge of a [[proton]].
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the '''specific charge''' of a [[Magnesium|Magnesium-24]] 2+ [[ion]].
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>Q = 1\times1.60\times10^{-19}</math>[[Coulomb|C]]
 +
 +
<math>m = 1\times1.67\times10^{-27}</math>[[Kilogram|kg]]
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
The [[Electrical Charge|charge]] on an [[Magnesium|Mg]] 2+ [[ion]] is 2 x [[Elementary Charge|elementary charge]].
 +
 +
<math>Q = 2\times1.60\times10^{-19}</math>[[Coulomb|C]]
 +
 +
The [[mass]] of [[Magnesium|Magnesium-24]] 2+ [[ion]] is the mass of 24 [[nucleon]]s + 10 [[electron]]s.
 +
 +
<math>m = (24\times1.67\times10^{-27} + 10\times9.11\times10^{-31})</math>[[Kilogram|kg]]
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>S.C. = \frac{1\times1.60\times10^{-19}}{1\times1.67\times10^{-27}}</math>
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>S.C. = \frac{2\times1.60\times10^{-19}}{24\times1.67\times10^{-27} + 10\times9.11\times10^{-31}}</math>
 +
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''3.  [[Solve (Maths)|solve]] the equation.'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>S.C. = \frac{1.60\times10^{-19}}{1.67\times10^{-27}}</math>
 +
 +
<math>S.C. = 9.58\times10^{7}Ckg^{-1}</math>
 +
 +
Correct to 3 [[Significant Figures|significant figures]].
 +
| style="height:20px; width:300px; text-align:left;" |'''3.  [[Solve (Maths)|solve]] the equation.'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>S.C. = \frac{3.20\times10^{-19}}{4.01\times10^{-26}}</math>
 +
 +
<math>S.C. = 7.98\times10^{6}Ckg^{-1}</math>
 +
 +
Correct to 3 [[Significant Figures|significant figures]].
 +
|}
 +
 
====Calculating Charge====
 
====Calculating Charge====
 +
{| class="wikitable"
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the [[Electrical Charge|charge]] of a [[Cobalt]] [[ion]] with a '''specific charge''' of 4.87x10<sup>6</sup>Ckg<sup>-1</sup> and a [[mass]] of 9.86x10<sup>-26</sup>kg.
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the [[Electrical Charge|charge]] of a [[Atomic Nucleus|nucleus]] with a '''specific charge''' of 4.51x10<sup>7</sup>Ckg<sup>-1</sup> and a [[mass]] of 8.52x10<sup>-26</sup>kg and identify the [[element]].
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>S.C._{Co} = 4.87\times10^{6}</math>Ckg<sup>-1</sup>
 +
 +
<math>m_{Co} = 9.86\times10^{-26}</math>[[Kilogram|kg]]
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>S.C. = 4.51\times10^{7}</math>Ckg<sup>-1</sup>
 +
 +
<math>m = 8.52\times10^{-26}</math>[[Kilogram|kg]]
 +
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>4.87\times10^{6} = \frac{Q}{9.86\times10^{-26}}</math>
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>4.51\times10^{7} = \frac{Q}{8.52\times10^{-26}}</math>
 +
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
 +
 +
<math>Q = 4.87\times10^{6} \times 9.86\times10^{-26}</math>
 +
 +
<math>Q = 4.80\times10^{-19}</math>[[Coulomb|C]]
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
 +
 +
<math>Q = 4.51\times10^{7} \times 8.52\times10^{-26}</math>
 +
 +
<math>Q = 3.84\times10^{-18}</math>[[Coulomb|C]]
 +
 +
To identify the [[Atomic Nucleus|nucleus]] divide the [[Electrical Charge|charge]] by the [[Elementary Charge|elementary charge]] to find the number of [[proton]]s.
 +
 +
<math>Z = \frac{3.84\times10^{-18}}{1.60\times10^{-19}}</math>
 +
 +
<math>Z = 24.0</math>
 +
 +
There are 24 [[proton]]s in the [[Atomic Nucleus|nucleus]] therefore the [[element]] is [[Chromium]].
 +
|}
 +
 
====Calculating Mass====
 
====Calculating Mass====
−
====Calculating Atomic Mass====
+
{| class="wikitable"
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the [[mass]] of an [[ion]] with a [[Electrical Charge|charge]] of -3.20x10<sup>-19</sup>[[Coulomb|C]] and a '''specific charge''' of -5.80x10<sup>6</sup>Ckg<sup>-1</sup>.
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the [[Relative Atomic Mass|relative atomic mass]] of a [[Atomic Nucleus|nucleus]] with an [[Electrical Charge|charge]] of 9.61x10<sup>-19</sup>[[Coulomb|C]] and a '''specific charge''' of 4.11x10<sup>7</sup>Ckg<sup>-1</sup>.
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 
 +
<math>S.C. = -5.80\times10^{6}</math>Ckg<sup>-1</sup>
 +
 
 +
<math>Q = -3.20\times10^{-19}</math>[[Coulomb|C]]
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 
 +
<math>S.C. = 4.11\times10^{7}</math>Ckg<sup>-1</sup>
 +
 
 +
<math>Q = 9.61\times10^{-19}</math>[[Coulomb|C]]
 +
|-
 +
 
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 
 +
<math>-5.80\times10^{6} = \frac{-3.20\times10^{-19}}{m}</math>
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 
 +
<math>4.11\times10^{7} = \frac{9.61\times10^{-19}}{m}</math>
 +
|-
 +
 
 +
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
 +
 
 +
<math>m=\frac{-3.20\times10^{-19}}{-5.80\times10^{6}}</math>
 +
 
 +
<math>m=5.52\times10^{-26}kg</math>
 +
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
 +
 
 +
<math>m=\frac{9.61\times10^{-19}}{4.11\times10^{7}}</math>
 +
 
 +
<math>m=2.34\times10^{-26}kg</math>
 +
 
 +
To find the [[Atomic Mass|atomic mass]] divide this [[mass]] by the [[mass]] of a single [[nucleon]].
 +
 
 +
<math>A=\frac{2.34\times10^{-26}}{1.67\times10^{-27}}</math>
 +
 
 +
<math>A=14.0</math>
 +
 
 +
The [[Relative Atomic Mass|relative atomic mass]] is 14.
 +
|}
 +
 
 
====Calculating the number of Neutrons====
 
====Calculating the number of Neutrons====
 +
{| class="wikitable"
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the number of [[neutron]]s in an [[isotope]] of [[Titanium]] with a [[Atomic Nucleus|nucleus]] of '''specific charge''' 4.58x10<sup>7</sup>Ckg<sup>-1</sup>.
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the number of [[neutron]]s in an [[isotope]] of [[Neon]] with the '''specific charge''' of -2.46x10<sup>6</sup>Ckg<sup>-1</sup> for a [[Selenium|Selenium 2-]] [[ion]].
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>S.C. = 4.58\times10^{7}</math>Ckg<sup>-1</sup>
 +
 +
There are 22 [[proton]]s in the [[Atomic Nucleus|nucleus]] of [[Titanium]].
 +
 +
<math>Q = 22\times1.60\times10^{-19}</math>C
 +
 +
<math>m = (z + n)\times1.67\times10^{-27}</math>kg
 +
 +
Where z is the number of [[proton]]s and n is the number of [[neutron]]s and (z+n) is the [[Relative Atomic Mass|relative atomic mass]].
 +
 +
Therefore
 +
 +
<math>m = (22 + n)\times1.67\times10^{-27}</math>kg
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>S.C. = -2.46\times10^{6}</math>Ckg<sup>-1</sup>
 +
 +
A 2- [[ion]] has two more [[electron]]s than [[proton]]s giving it a [[Negative Charge|negative charge]].
 +
 +
<math>Q = -2\times1.60\times10^{-19}</math>C
 +
 +
<math>m = (z + n)\times1.67\times10^{-27}+(z+2)\times9.11\times10^{-31}</math>kg
 +
 +
Where z is the number of [[proton]]s, n is the number of [[neutron]]s, (z+n) is the [[Relative Atomic Mass|relative atomic mass]] and (z+2) is the number of [[electron]]s since there are two more [[electron]]s than [[proton]]s.
 +
 +
Therefore
 +
 +
<math>m = (34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31}</math>kg
 +
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<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]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>-2.46\times10^{6} = \frac{-2\times1.60\times10^{-19}}{(34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31}}</math>
 +
 +
|-
 +
| 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\times1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}}</math>
 +
 +
<math>n = \frac{22\times1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}} - 22</math>
 +
 +
<math>n = 46.0 - 22</math>
 +
 +
<math>n = 24.0</math>
 +
 +
There are 24 [[neutron]]s in this isotope of [[Titanium]].
 +
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] and [[Evaluate (Maths)|evaluate]] the equation, then [[Solve (Maths)|solve]].'''
 +
 +
<math>(34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31} = \frac{-2\times1.60\times10^{-19}}{-2.46\times10^{6}}</math>
 +
 +
<math>(34 + n)\times1.67\times10^{-27}+3.2796\times10^{-29} = 1.3008\times10^{-25}</math>
 +
 +
<math>(34 + n) = \frac{1.3008\times10^{-25}-3.2796\times10^{-29}}{1.67\times10^{-27}}</math>
 +
 +
<math>(34 + n) = 77.9</math>
 +
 +
<math>n = 43.9</math>
 +
The [[Atomic Nucleus|nucleus]] contains 44 [[neutron]]s.
 +
|}
 +
 
====Calculating Atomic Number====
 
====Calculating Atomic Number====
 +
{| class="wikitable"
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the number of [[proton]]s in a [[Atomic Nucleus|atomic nucleus]] of an [[isotope]] with an [[Relative Atomic Mass|atomic mass]] of 231[[Unified Atomic Mass Units|u]] and a [[Specific Charge|specific charge]] of 3.82x10<sup>7</sup>Ckg<sup>-1</sup>.
 +
| style="height:20px; width:300px; text-align:center;" |Calculate the number of [[proton]]s in a [[Atomic Nucleus|atomic nucleus]] of an [[isotope]] with 30 [[neutron]]s and a [[Specific Charge|specific charge]] of 4.35x10<sup>7</sup>Ckg<sup>-1</sup>.
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>S.C. = 3.82\times10^{7}</math>Ckg<sup>-1</sup>
 +
 +
<math>Q = z\times1.60\times10^{-19}</math>C
 +
 +
<math>m = 231\times1.67\times10^{-27}</math>kg
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>S.C. = 4.35\times10^{7}</math>Ckg<sup>-1</sup>
 +
 +
<math>Q = z\times1.60\times10^{-19}</math>C
 +
 +
The [[Relative Atomic Mass|atomic mass]] can be approximated as the number of [[nucleon]]s multiplied by the [[Unified Atomic Mass Units|unified atomic mass unit]].
 +
 +
<math>m = (z+30)\times1.67\times10^{-27}</math>kg
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>3.82\times10^{7} = \frac{z\times1.60\times10^{-19}}{231\times1.67\times10^{-27}}</math>
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers and [[Evaluate (Maths)|evaluate]].'''
 +
 +
<math>S.C. = \frac{Q}{m}</math>
 +
 +
<math>4.35\times10^{7} = \frac{z\times1.60\times10^{-19}}{(z+30)\times1.67\times10^{-27}}</math>
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
 +
 +
<math>3.82\times10^{7}\times231\times1.67\times10^{-27} = z\times1.60\times10^{-19}</math>
 +
 +
<math>1.4736414\times10^{-17} = z\times1.60\times10^{-19}</math>
 +
 +
<math>z = \frac{1.4736414\times10^{-17}}{1.60\times10^{-19}}</math>
 +
 +
<math>z = 92.1</math>
 +
 +
The [[element]] is [[Uranium]].
 +
| style="height:20px; width:300px; text-align:left;" |'''3. [[Rearrange (Maths)|Rearrange]] the equation and [[Solve (Maths)|solve]].'''
 +
 +
<math>4.35\times10^{7} = \frac{z\times1.60\times10^{-19}}{(z+30)\times1.67\times10^{-27}}</math>
 +
 +
<math>4.35\times10^{7}\times(z+30)\times1.67\times10^{-27} = z\times1.60\times10^{-19}</math>
 +
 +
<math>(z+30)\times7.2645\times10^{-20} = z\times1.60\times10^{-19}</math>
 +
 +
<math>z\times7.2645\times10^{-20} + 30\times7.2645\times10^{-20}= z\times1.60\times10^{-19}</math>
 +
 +
<math>z\times1.60\times10^{-19} - z\times7.2645\times10^{-20} = 30\times7.2645\times10^{-20}</math>
 +
 +
<math>z(1.60\times10^{-19} - 7.2645\times10^{-20}) = 30\times7.2645\times10^{-20}</math>
 +
 +
<math>z = \frac{30\times7.2645\times10^{-20}}{1.60\times10^{-19} - 7.2645\times10^{-20}}</math>
 +
 +
<math>z = 24.9</math>
 +
 +
There are 25 [[proton]]s so the [[element]] is [[Manganese]].
 +
|}

Latest revision as of 12:16, 29 July 2019

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.

\(Q = 1\times1.60\times10^{-19}\)C

\(m = 1\times1.67\times10^{-27}\)kg

1. State the known quantities in SI Units.

The charge on an Mg 2+ ion is 2 x elementary charge.

\(Q = 2\times1.60\times10^{-19}\)C

The mass of Magnesium-24 2+ ion is the mass of 24 nucleons + 10 electrons.

\(m = (24\times1.67\times10^{-27} + 10\times9.11\times10^{-31})\)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 a 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 -2.46x106Ckg-1 for a Selenium 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.

\(S.C. = -2.46\times10^{6}\)Ckg-1

A 2- ion has two more electrons than protons giving it a negative charge.

\(Q = -2\times1.60\times10^{-19}\)C

\(m = (z + n)\times1.67\times10^{-27}+(z+2)\times9.11\times10^{-31}\)kg

Where z is the number of protons, n is the number of neutrons, (z+n) is the relative atomic mass and (z+2) is the number of electrons since there are two more electrons than protons.

Therefore

\(m = (34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31}\)kg

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.

\(S.C. = \frac{Q}{m}\)

\(-2.46\times10^{6} = \frac{-2\times1.60\times10^{-19}}{(34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31}}\)

3. Rearrange and evaluate the equation, then solve.

\((22 + n) = \frac{22\times1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}}\)

\(n = \frac{22\times1.60\times10^{-19}}{1.67\times10^{-27}\times4.58\times10^{7}} - 22\)

\(n = 46.0 - 22\)

\(n = 24.0\)

There are 24 neutrons in this isotope of Titanium.

3. Rearrange and evaluate the equation, then solve.

\((34 + n)\times1.67\times10^{-27}+36\times9.11\times10^{-31} = \frac{-2\times1.60\times10^{-19}}{-2.46\times10^{6}}\)

\((34 + n)\times1.67\times10^{-27}+3.2796\times10^{-29} = 1.3008\times10^{-25}\)

\((34 + n) = \frac{1.3008\times10^{-25}-3.2796\times10^{-29}}{1.67\times10^{-27}}\)

\((34 + n) = 77.9\)

\(n = 43.9\) The nucleus contains 44 neutrons.

Calculating Atomic Number

Calculate the number of protons in a atomic nucleus of an isotope with an atomic mass of 231u and a specific charge of 3.82x107Ckg-1. Calculate the number of protons in a atomic nucleus of an isotope with 30 neutrons and a specific charge of 4.35x107Ckg-1.
1. State the known quantities in SI Units.

\(S.C. = 3.82\times10^{7}\)Ckg-1

\(Q = z\times1.60\times10^{-19}\)C

\(m = 231\times1.67\times10^{-27}\)kg

1. State the known quantities in SI Units.

\(S.C. = 4.35\times10^{7}\)Ckg-1

\(Q = z\times1.60\times10^{-19}\)C

The atomic mass can be approximated as the number of nucleons multiplied by the unified atomic mass unit.

\(m = (z+30)\times1.67\times10^{-27}\)kg

2. Substitute the numbers and evaluate.

\(S.C. = \frac{Q}{m}\)

\(3.82\times10^{7} = \frac{z\times1.60\times10^{-19}}{231\times1.67\times10^{-27}}\)

2. Substitute the numbers and evaluate.

\(S.C. = \frac{Q}{m}\)

\(4.35\times10^{7} = \frac{z\times1.60\times10^{-19}}{(z+30)\times1.67\times10^{-27}}\)

3. Rearrange the equation and solve.

\(3.82\times10^{7}\times231\times1.67\times10^{-27} = z\times1.60\times10^{-19}\)

\(1.4736414\times10^{-17} = z\times1.60\times10^{-19}\)

\(z = \frac{1.4736414\times10^{-17}}{1.60\times10^{-19}}\)

\(z = 92.1\)

The element is Uranium.

3. Rearrange the equation and solve.

\(4.35\times10^{7} = \frac{z\times1.60\times10^{-19}}{(z+30)\times1.67\times10^{-27}}\)

\(4.35\times10^{7}\times(z+30)\times1.67\times10^{-27} = z\times1.60\times10^{-19}\)

\((z+30)\times7.2645\times10^{-20} = z\times1.60\times10^{-19}\)

\(z\times7.2645\times10^{-20} + 30\times7.2645\times10^{-20}= z\times1.60\times10^{-19}\)

\(z\times1.60\times10^{-19} - z\times7.2645\times10^{-20} = 30\times7.2645\times10^{-20}\)

\(z(1.60\times10^{-19} - 7.2645\times10^{-20}) = 30\times7.2645\times10^{-20}\)

\(z = \frac{30\times7.2645\times10^{-20}}{1.60\times10^{-19} - 7.2645\times10^{-20}}\)

\(z = 24.9\)

There are 25 protons so the element is Manganese.

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