Skip to main content
Ch.2 - Atoms, Molecules, and Ions

Chapter 2, Problem 88c

The natural abundance of 3He is 0.000137%. (c) Based on your answer to part (b), what would need to be the precision of a mass spectrometer that is able to differentiate between peaks that are due to 3He+ and 3H+?

Verified Solution
Video duration:
6m
This video solution was recommended by our tutors as helpful for the problem above.
565
views
Was this helpful?

Video transcript

Hi everyone here we have a question telling us that phosphorus 32 is a radioactive isotope of phosphorus and undergoes a beta decay to yield Sulfur 32. If the two isotopes are to be differentiated using a mass spectrometer, what should be the precision of the mass spectrometer spectrometer that is able to differentiate between the two. So to determine the precision of the mass spectrometer. That is able to differentiate between the two. We first need to know the masses and to find the masses of phosphorus 32 and Sulfur 32. We need to know the masses of a proton and neutron because that is what constitutes the mass of an atom. So the mass of a proton Is 1. two times 10 To the -24 g. The mass of a neutron Is 1. 721, 1 times 10 to the negative 24 grams. So now we can calculate the mass of phosphorus 32. So its atomic number. The number on the periodic table is 15, Which means there are 15 protons. And now we need to find how many neutrons we have. So the neutrons equals the mass number minus the atomic number, Which in this case the mass number is 32. That was given to us Minour the atomic number which is on the period of table, which is 15 And that equals 17. So our mass of 32 phosphorous equals 15 times 1.67 to to 1673 times 10 to the negative 24th plus 17 times 1. times 10 to the negative 24th, which equals 5. times 10 to the negative 23rd grams. And now we're going to do the exact same thing for Sulfur 32. The atomic number is the number of protons again, which is 16. The neutrons Is the mass number minus the atomic number. So in this case it's -16 which equals 16. So our mass of 32 sulfur equals 16 Times 1. - six. 21673 times 10 to the negative 24th plus 16 times 1. times 10 to the negative 24th And that equals 5. times 10 to the negative 23 g. And now to find the precision that the mass spectrum chair needs to be, we need to find the difference between the two masses. So 5. times 10 to the negative 23 g minus 5. times 10. To the negative 23 g Equals 2. times 10 to the -27 g. So we need the mass spectrometer to have a precision of at least. 2.306 times 10 to the - g. And that is our final answer. Thank you for watching. Bye.
Related Practice
Textbook Question

Suppose a scientist repeats the Millikan oil-drop experiment but reports the charges on the drops using an unusual (and imaginary) unit called the warmomb (wa). The scientist obtains the following data for four of the drops: Droplet Calculated Charge (wa) A 3.84⨉10−8 B 4.80⨉10−8 C 2.88⨉10−8 D 8.64⨉10−8 (d) What is the conversion factor between warmombs and coulombs?

1455
views
Textbook Question

The natural abundance of 3He is 0.000137%. (a) How many protons, neutrons, and electrons are in an atom of 3He?

806
views
Textbook Question

The natural abundance of 3He is 0.000137%. (b) Based on the sum of the masses of their subatomic particles, which is expected to be more massive, an atom of 3He or an atom of 3H (which is also called tritium)?

906
views
Textbook Question

A cube of gold that is 1.00 cm on a side has a mass of 19.3 g. A single gold atom has a mass of 197.0 u. (a) How many gold atoms are in the cube?

1045
views
Textbook Question

A cube of gold that is 1.00 cm on a side has a mass of 19.3 g. A single gold atom has a mass of 197.0 u. (b) From the information given, estimate the diameter in Å of a single gold atom.

912
views
Textbook Question

The diameter of a rubidium atom is 495 pm We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is 1.0 cm on a side?

488
views