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→Calculating Atomic Number
====Calculating Specific Charge====
{| class="wikitable"
| style="height:20px; width:200px300px; text-align:center;" |QuestionCalculate the '''specific charge of a [[proton]].| style="height:20px; width:200px300px; text-align:center;" |QuestionCalculate the '''specific charge''' of a [[Magnesium|Magnesium-24]] 2+ [[ion]].
|-
| style="height:20px; width:200px300px; 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:200px300px; 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:200px300px; 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:200px300px; 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:200px300px; text-align:left;" |'''3. [[Rearrange Solve (Maths)|Rearrangesolve]] the equation and .''' <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 [[Solve (Maths)Significant Figures|solvesignificant figures]].'''| style="height:20px; width:200px300px; text-align:left;" |'''3. [[Rearrange Solve (Maths)|Rearrangesolve]] the equation and .''' <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 [[Solve (Maths)Significant Figures|solvesignificant figures]].'''
|}
====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====
{| 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==Calculating "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====
{| 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====
{| 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]].
|}