Specific Charge

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.