A muon has a rest mass of 1.89 times 10 to the negative 28 kilograms. If the muon is moving at a speed of 20 meters per second, what is its de Broglie wavelength? Use a value of 6.63 times 10 to the negative 34th joules-seconds for the Planck constant. Give your answer in scientific notation to two decimal places.
The question asks us to find a de Broglie wavelength, which means we will need the de Broglie wavelength relation. In particular, we have that the wavelength of any particle is equal to the Planck constant divided by that particle’s momentum. This relationship is true for massless particles and also for massive particles regardless of how fast they’re moving. We only need to be careful that for massive particles with speeds close to the speed of light, we use the relativistic momentum.
However, our muon is moving at only 20 meters per second, which is seven orders of magnitude smaller than the speed of light. We can therefore calculate the momentum as mass times velocity and use this nonrelativistic momentum in the de Broglie relation. We are told that the rest mass of our muon is 1.89 times 10 to the negative 28 kilograms. And again, because the muon is moving at nonrelativistic speeds, this is the mass that we will use in our formula. Multiplying 1.89 times 10 to the negative 28 kilograms by 20 meters per second gives us 3.78 times 10 to the negative 27 in units of kilogram-meters per second. When we substitute this value along with 6.63 times 10 to the negative 34 joule-seconds for the Planck constant into the de Broglie wavelength formula, we get 6.63 times 10 to the negative 34 joule-seconds divided by 3.78 times 10 to the negative 27 kilogram-meters per second.
Let’s start this calculation by looking at the units. Joules are equivalent to kilograms meters squared per second squared. So joule-seconds are kilograms meters squared per second. So our final units are kilograms meters squared per second divided by kilogram-meters per second. The only difference between these two sets of units is the extra factor of meters on top. So when we remove like units from the numerator and denominator, we are left with units of just meters. This tells us we’re on the right track because we’re looking for a wavelength. And meters are one possible unit to use when reporting a length.
Let’s next look at the powers of 10. The numerator and denominator both have the same base, namely, 10, raised to different exponents. The result will be that same base raised to the difference between the exponents, here negative 34 minus negative 27. Negative 34 minus negative 27 is negative seven. So the resulting power of 10 is 10 to the negative seven.
Finally, we need to calculate 6.63 divided by 3.78. Plugging this into a calculator gives 1.7539 and several more decimal places. However, the question only asks us for the calculation to two decimal places. So, to round to two decimal places, we take a look at the third decimal place, which is three. Since this is less than five, we don’t change the number at all. And 1.7539 et cetera rounded to two decimal places is simply 1.75.
And so, to two decimal places in scientific notation, the de Broglie wavelength of our muon is 1.75 times 10 to the negative seven meters.