The atomic masses of 203tl and 205tl are 202.972336 and 204.974410 amu, respectively. the average atomic mass of thallium is 204.383 amu. calculate the natural abundances of these two isotopes.

Answer : Natural abundance of Tl-203 is 29.54% and that of Tl-205 is 70.46%.

Explanation :

The total natural abundance for all the isotopes of the element is always 100% or 1 .

Let us assume natural abundance of Tl-203 is a.

Therefore natural abundance of Tl-205 would be (1-a)

The average atomic mass of an element is the total of natural abundance multiplied by its isotopic mass.

The formula for average atomic mass can be written as,

Average Atomic Mass = Mass of Tl-203 x (a) + Mass of Tl-205 x (1-a)

Average Atomic Mass =

Average atomic mass of Tl is 204.383 amu. Let us plug in this value in the above formula.

In percent form, this can be expressed as 29.54%

percent abundance of Tl-203 is 29.54%

Percent abundance of Tl-205 is 100 - 29.54 = 70.46%

The natural abundances of these two isotopes are 29.54%  and 70.46%

The elements in nature have several types of isotopes

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Atomic mass is the average atomic mass of all its isotopes

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu

So the atomic mass obtained is the mass of the atom relative to the 12th carbon atom

An atomic mass unit = amu is a relative atomic mass of 1/12 the mass of an atom of carbon-12.

The 'amu' unit has now been replaced with a unit of 'u' only

for example, Carbon has 3 isotopes, namely ₆¹²C, ₆¹³C, and ₆¹⁴C

Mass atom X = mass isotope 1 . % + mass isotope 2.%

The atomic masses of 203tl and 205tl  has two isotopes, 202.972336 and 204.974410 amu,

The average atomic mass  is 204.383 amu.

For example isotopes 202.972336 = x% and isotopes 204.974410 amu = 100-x%

Mass atom element  = mass isotope 1 . % + mass isotope 2.%

204.383 amu= 202.972336  %x +204.974410. (100-x)%

204.383 amu = 202.972336%x + 204.974410 .100% - 204.974410 %x

204.383 amu = -2.002074 %x + 204.974410

-0.59141 amu = -2.002074%x  

%x = 0.2954

x = 29.54%

So percent abundance the other isotopes = 100% - 29.54% = 70.46%

The subatomic particle that has the least mass

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Keywords: mass number, atomic mass, amu, isotope

5.6⋅10−^19  j

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