Applications of Scientific Notation
Compare cosmic and atomic numbers, multiply and divide in scientific notation, round to significant figures, and use the EE/EXP key on your calculator.
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The mass of the Earth is about $6 \times 10^{24}$ kg. The mass of a person is about $6 \times 10^{1}$ kg. About how many people would weigh the same as the Earth? Estimate the index first — then check the coefficient.
Scientific notation lets us multiply and divide colossal or tiny numbers using just the index laws — coefficients use ordinary arithmetic, and the powers of 10 use the product and quotient rules.
$(a \times 10^m) \times (b \times 10^n) = (a \times b) \times 10^{m+n}$ — multiply coefficients, add indices. $\dfrac{a \times 10^m}{b \times 10^n} = \dfrac{a}{b} \times 10^{m-n}$ — divide coefficients, subtract indices. Adjust at the end so the coefficient sits between $1$ and $10$.
Know
- Standard form $a \times 10^n$ where $1 \le a < 10$
- Multiply / divide rules in scientific notation
- Meaning of significant figures (sig fig)
Understand
- Why scientific notation makes huge / tiny numbers comparable
- How to re-adjust a coefficient that drifts out of $[1, 10)$
- Why 3 sig fig is the usual answer precision in science
Can Do
- Compute $(3 \times 10^8) \times (2 \times 10^5)$
- Compare $1.5 \times 10^{11}$ to $5 \times 10^{-11}$
- Use the EE / EXP key on a scientific calculator
Wrong: “$(4 \times 10^5) \times (3 \times 10^6) = 12 \times 10^{11}$” left as-is — the coefficient $12$ is not in $[1, 10)$.
Right: $12 \times 10^{11} = 1.2 \times 10^{12}$. Tidy the coefficient.
Wrong: Typing “$3 \times 10$ EXP $8$” on the calculator — you've entered $3 \times 10 \times 10^8 = 3 \times 10^9$.
Right: Type “$3$ EXP $8$”. The EXP key is the $\times 10^?$ part already.
Compare the powers of 10 first. Whichever has the larger index is bigger (for positive coefficients). Only if the indices tie do you compare the coefficients.
Which is bigger: $7 \times 10^9$ or $2 \times 10^{10}$? The indices are $9$ and $10$, so $2 \times 10^{10}$ wins — even though its coefficient is smaller. How many times bigger? $\dfrac{2 \times 10^{10}}{7 \times 10^9} = \dfrac{2}{7} \times 10^{1} \approx 2.86$ times bigger.
Sig figs count the digits that carry information. Leading zeros never count; trailing zeros after a decimal point do count. Most science answers are quoted to 3 sig fig.
$0.00420$ has $3$ sig fig (the digits $4$, $2$, $0$). $5{,}730{,}000$ written as $5.73 \times 10^6$ shows $3$ sig fig clearly. Round $2.4763 \times 10^8$ to $3$ sig fig: look at the 4th digit ($6 \ge 5$), so round up — $2.48 \times 10^8$.
Watch Me Solve It · 3 examples
- 1Multiply the coefficients$4 \times 3 = 12$
- 2Add the indices (product rule on powers of 10)$10^5 \times 10^6 = 10^{11}$
- 3Re-normalise so $1 \le a < 10$$12 \times 10^{11} = 1.2 \times 10^{12}$Decimal shifts one left $\to$ index goes up by 1.
- 1Set up as a quotient$N = \dfrac{1.5 \times 10^{11}}{5 \times 10^{-11}}$
- 2Divide coefficients, subtract indices$\dfrac{1.5}{5} = 0.3$; $10^{11 - (-11)} = 10^{22}$
- 3Re-normalise$0.3 \times 10^{22} = 3 \times 10^{21}$Coefficient shifts right by one $\to$ index drops by 1.
- 1Write the quotient$N = \dfrac{1}{9.5 \times 10^{-13}}$
- 2Divide $1$ by the coefficient$\dfrac{1}{9.5} \approx 0.10526$
- 3Flip the index sign (dividing by $10^{-13}$ = $\times 10^{13}$) and re-normalise to 3 s.f.$0.10526 \times 10^{13} = 1.0526 \times 10^{12} \approx 1.05 \times 10^{12}$
Common Pitfalls
Multiply
- $(a \times 10^m)(b \times 10^n) = ab \times 10^{m+n}$
- Re-normalise after
Divide
- $\dfrac{a \times 10^m}{b \times 10^n} = \dfrac{a}{b} \times 10^{m-n}$
- Subtract indices carefully with signs
Compare
- Bigger index = bigger number
- Same index $\to$ compare coefficients
Significant figures
- 3 s.f. is standard in science
- Round the digit after the last kept digit
How are you completing this lesson?
Brain Trainer · 4 problems
Mix multiplication, division and a real-world question. Round answers to 3 sig fig when asked.
1 Calculate $(2 \times 10^7) \times (4 \times 10^3)$.
$2 \times 4 = 8$; $10^{7+3} = 10^{10}$.$8 \times 10^{10}$2 Calculate $\dfrac{8 \times 10^{12}}{2 \times 10^{-4}}$.
$8 / 2 = 4$; $10^{12 - (-4)} = 10^{16}$.$4 \times 10^{16}$3 The world's population is about $8.0 \times 10^9$. If everyone held $1.5$ m of arm-span, what total length is that, to 2 s.f.?
$8.0 \times 1.5 = 12.0$; $\times 10^9$ m $= 1.2 \times 10^{10}$ m.$1.2 \times 10^{10}$ m4 Round $4.6738 \times 10^{-5}$ to 3 sig fig.
Keep first 3 digits: $4$, $6$, $7$. Next digit $3 < 5$, so round down.$4.67 \times 10^{-5}$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Calculate, leaving the answer in scientific notation: (a) $(6 \times 10^4) \times (5 \times 10^7)$, (b) $\dfrac{4.8 \times 10^{-3}}{1.6 \times 10^{2}}$, (c) $(2 \times 10^5)^3$.
Q7. The mass of an electron is $9.11 \times 10^{-31}$ kg. The mass of a proton is $1.67 \times 10^{-27}$ kg. (a) Which is heavier? (b) How many times heavier, to 3 sig fig?
Q8. Australia's population is about $2.6 \times 10^7$. The federal budget is about $\$6.5 \times 10^{11}$. Calculate the spending per person, to 3 sig fig. Show your calculator key-sequence using EE / EXP.
Quick Check
1. B — $2 \times 10^{10}$.
2. C — $3 \times 10^{10}$.
3. A — $2 \times 10^{10}$.
4. D — $3.05 \times 10^7$.
5. B — $5 \times 10^{2}$ s ($\approx 500$ s, about 8.3 minutes).
Show Your Working Model Answers
Q6 (3 marks): (a) $6 \times 5 = 30$, $10^{4+7} = 10^{11}$, re-normalise: $3 \times 10^{12}$ [1]; (b) $4.8/1.6 = 3$, $10^{-3-2} = 10^{-5}$, so $3 \times 10^{-5}$ [1]; (c) $2^3 = 8$, $10^{5 \times 3} = 10^{15}$, so $8 \times 10^{15}$ [1].
Q7 (3 marks): (a) Indices: $-27 > -31$, so the proton is heavier [1]. (b) Ratio = $\dfrac{1.67}{9.11} \times 10^{-27 - (-31)} = 0.1833 \times 10^{4} = 1.833 \times 10^{3}$ [1], i.e. about $1{,}830$ times heavier ($\approx 1.83 \times 10^3$ to 3 s.f.) [1].
Q8 (3 marks): Keys: $6.5$ EXP $11$ $\div$ $2.6$ EXP $7$ $=$ [1]. Coefficient: $6.5 / 2.6 = 2.500$ [1]. Index: $10^{11-7} = 10^4$. Answer: $2.50 \times 10^4 = \$25{,}000$ per person (3 s.f.) [1].
Atomic Headcount
A single grain of table salt contains roughly $6.0 \times 10^{18}$ formula units of NaCl. If you eat $5.0$ g of salt at dinner (one grain $\approx 5.85 \times 10^{-5}$ g), about how many formula units have you swallowed? Give your answer in scientific notation to 2 sig fig, then state the order of magnitude.
Reveal solution
Grains: $\dfrac{5.0}{5.85 \times 10^{-5}} \approx 8.547 \times 10^{4}$. Formula units: $8.547 \times 10^{4} \times 6.0 \times 10^{18} \approx 51.28 \times 10^{22} = 5.1 \times 10^{23}$. Order of magnitude $\approx 10^{23}$.
Multiply
$ab \times 10^{m+n}$
Divide
$\dfrac{a}{b} \times 10^{m-n}$
Compare
Index first, then coefficient
Re-normalise
Keep $1 \le a < 10$
3 sig fig
Standard for science
EE / EXP
Replaces $\times 10^?$
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