A particle has a de Broglie wavelength of 0.200 nanometers. What is its momentum? Use a value of 6.63 times 10 to the negative 34 joule-seconds for the Planck constant. Give your answer in scientific notation to two decimal places.
Our first step to answer this question is to recall that the de Broglie wavelength of a particle and the particle’s momentum are related by the formula 𝜆 equals ℎ divided by 𝑝, where 𝜆 is the de Broglie wavelength of the particle, ℎ is the Planck constant, and 𝑝 is the particle’s momentum. We are given values for the wavelength and the Planck constant. So we just need to rearrange this formula to solve for the momentum.
If we multiply both sides by momentum and divide both sides by wavelength, on the left we are left with 𝑝 and on the right we are left with ℎ divided by 𝜆. All that’s left now is to substitute values. Before that though, let’s rewrite the wavelength in a more convenient form. Our final answer is meant to be expressed in scientific notation. And also the wavelength is expressed in terms of nanometers, which are not SI base units. So let’s convert 0.200 nanometers to a number in scientific notation with SI base units. One nanometer is 10 to the negative nine meters. So the wavelength of our particle is 0.200 times 10 to the negative nine meters.
Now, this number is not quite in scientific notation because the leading digit is zero instead of a number greater than zero. To fix this, we need to move the decimal place over so that it is after the first nonzero digit. For every place that we move the decimal point to the right, we will need to subtract one from the exponent. This is because moving the decimal point to the right is equivalent to multiplying by 10. And subtracting one from the exponent is equivalent to dividing by 10. So the overall result is multiplying by 10 and dividing by 10, which leaves the actual value unchanged.
So, in scientific notation using SI base units, the wavelength is two times 10 to the negative 10 meters. Substituting into our formula, we have that the momentum is 6.63 times 10 to the negative 34 joule-seconds divided by 2.00 times 10 to the negative 10 meters. At this point, we can check if we are on the right track by paying attention to our units, joule-seconds divided by meters. Since joules are kilograms meters squared per second squared, our units are kilograms meters squared per second squared times seconds per meter. Meters squared per meter is just meters, and seconds per second squared is just per second. So this simplifies to kilograms meters per second. Kilograms are units of mass, and meters per second are units of velocity. And momentum is mass times velocity. So kilograms meters per second are the right units for what we are looking for.
For the numerical portion of this calculation, we have 6.63 divided by 2.00, which is 3.315. And for the power of 10, 10 to the negative 34 divided by 10 to the negative 10 is 10 to the negative 34 plus 10, which is 10 to the negative 24. Combining our numerical result with the units, we get 3.315 times 10 to the negative 24 kilogram-meters per second.
All that’s left now is to round our answer to two decimal places. The digit in the third decimal place is five. So, to round to two decimal places, we increment the second decimal place by one, which gives us 3.32. And our final answer written in scientific notation to two decimal places is 3.32 times 10 to the negative 24 kilogram-meters per second.