Scientific Notation

Scientific notation is a system of writing numbers so that they are easily evaluated and manipulated. The format is standardized to be easy to read, with one single-digit integer written ahead of the decimal point and an exponent showing the overall magnitude of the number.

Scientific notation offers a way to avoid the tedium, awkwardness, and potential for error that comes with calculations involving very large or very small numbers. How much does the sun weigh? How much does a single hydrogen atom weigh? Questions from astronomy, physics, chemistry, geology, and biology often involve scales that are difficult to conceptualize and manipulate.

Scientific notation, shown on the left-hand side of this equation, eliminates placeholder zeroes, making numbers easier to use in calculations.

The sun, a star made up mostly of hydrogen and helium, weighs roughly 2,000,000,000,000,000,000,000,000,000,000,000 grams.

Toward the other end of the spectrum, a single hydrogen atom weighs around 0.0000000000000000000000017 grams.

Very few calculators have displays that reach 24 decimal places.

The values above can also be expressed as $2 \times 10^{33}\text{ g}$ for the weight of the sun and $1.7 \times 10^{-24}\text{ g}$ for the hydrogen atom, numbers that are far easier to manipulate mathematically and perhaps easier to understand as well. Since people do not have any hands-on experience with objects as tiny as an atom or as heavy as a star, these values can be hard to comprehend, let alone compare, without exponents. Scientific notation shows the relationship between different values and allows for easier movement between scales.

Scientific notation is written using 10 as the base number and an exponent, as in $a \times 10^b,$ where

• $1 \leq a < 10$
• $b$ is an integer, which can be positive or negative.

A simple way to convert a number to scientific notation is to move the decimal point to the left or right the number of places needed to get a number $a$ such that $1 \leq a < 10$. The number of places the decimal point was moved is $b$. If the decimal point was moved to the left, $b$ is positive $($the original number is larger than $b)$ and if the decimal point was moved to the right, $b$ is negative and the number being represented is smaller than $b$.

Write $1234567$ in scientific notation.

ANSWER

We have

$$1.234567 \times 10^{6}.\ _\square$$

Write $0.000000901$ in scientific notation.

ANSWER

We have

$$0.000000901 = 9.01 \times 10^{-7}.\ _\square$$

Scientific notation is helpful in two ways:

1. The exponents allow for quick and accurate comparisons. It is much easier to evaluate which number is larger between $3 \times 10^{8}$ and $3 \times 10^{11}$ than it is to compare $300000000$ and $300000000000$.
2. It conveys the number of significant figures in a measurement. Thus, $1.2000 \times 10^{-5}$ tells that the measurement is known with more certainty than $1.2 \times 10^{-5}$ even though they are numerically equivalent.

Numbers written in scientific notation cannot be added or subtracted unless their exponents $b$ all have the same value. The operations are then carried out on the values of $a$.

Add $1.42 \times 10^2$ and $3.45 \times 10^3.$

ANSWER

Both exponents must have the same value. Choosing either option will give the same answer:

$$\begin{aligned} 3.45 \times 10^3 &= 34.5 \times 10^2\\ \Rightarrow 34.5 \times 10^2 + 1.42 \times 10^2 &= 35.92 \times 10^2. \end{aligned}$$

The convention of scientific notation requires a number less than 10, so the answer can be rewritten as $3.592\times 10^3.$

Similarly,

$$\begin{aligned} 1.42 \times 10^2 &= 0.142 \times 10^3\\ \Rightarrow 0.142 \times 10^3 + 3.45 \times 10^3 &= 3.592 \times 10^3.\ _\square \end{aligned}$$

For multiplication, multiply the values of $a$ and add the values of $b:$

$$(C\times 10^{c}) \times (D\times 10^{d}) = CD\times 10^{(c+d)}.$$

For division, divide the dividend's value of $a$ by the divisor's value of $a,$ and then subtract the exponent $b$ of the divisor from $b$ of the dividend:

$$(F\times 10^{f}) \div (H\times 10^{h}) = (F \div H) \times 10^{(f-h)}.$$

Often, the answer obtained by following this process will not be in standard scientific notation, so a final step (moving the decimal point and adjusting the exponent) is needed to convert the answer to the proper format.

Write the answer to the following problem in scientific notation:

$$\left[3.8\times 10^{4}\right] \times \left[9.6\times 10^{2}\right].$$

ANSWER

We have

$$\begin{aligned} \left[3.8\times 10^{4}\right] \times \left[9.6\times 10^{2}\right] &= [3.8\times 9.6] \times 10^{(4+2)}\\ &= [36.48] \times 10^{6}\\ &= 3.648 \times 10^{7}.\ _\square \end{aligned}$$

An atom is mostly empty space. That's a common sentence in general chemistry textbooks that seems fairly straightforward. Then, the diagrams in the textbook look something like this:

Diagrams of atoms are usually not drawn to scale. Image from Wikimedia Commons user Rainer Klute.

Based on this image, a student might reasonably believe that the nucleus takes up about 10% of the atom's space, while 90% of the atom is empty space. However, the diameters of the nucleus and the electron orbital are not drawn to scale. Looking at those numerical values paints a wildly different picture from the one shown above.

The best estimate of a hydrogen atom's diameter is about 1.7566 femtometers. A femtometer is $10^{-15} ~\si{\meter}$. Rewrite the diameter in meters using scientific notation.

ANSWER

We have

$$1.7566\times10^{-15} ~\si{\meter}.\ _\square$$

The diameter of a hydrogen atom's orbiting electron can be estimated using Bohr's model and is approximately $5.29177\times10^{-11} ~\si{\meter}$. So the diameter of the electron orbit is $\big( 5.29177\times10^{-11} \big) \div \big( 1.7566\times10^{-15} \big) \approx 30000$ times greater than the diameter of the proton. That difference in scale can be used to make a macroscopic analogy:

If a proton was the size of a tennis ball, how big would the electron orbital be on a hydrogen atom?

Poll a group of general chemistry students, and their answers may vary widely. The size of a soccer ball? The size of a soccer stadium? The size of the city? The estimates may depend on what images the student has seen in textbooks, but many of them will be off by several orders of magnitude.

A tennis ball is about $6.5 \si{\centi\meter},$ or $6.5\times10^{-2}\ \si{ \meter}$, in diameter. The electron orbit's diameter would be 30000 times bigger than the tennis ball's, or $6.5 \times 10^{-2}\ \si{ \meter} \times 30000 \approx 1950\ \si{ \meter}.$ A tennis court is about 24 meters long. So the electron orbit's diameter is $1950\ \si{ \meter}\div 24\ \si{ \meter}\approx 81$ tennis courts long.