# Boron Has Two Stable Isotopes, 105b510b and 115b511b. Their Respective Masses Are 10.01294 U and 11.00931 U, and the Atomic Mass of Boron is 10.811 U. Find the Abundances of 105b510b and 115b511b. - Physics

Numerical

Boron has two stable isotopes, ""_5^10"B" and ""_5^11"B". Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of  ""_5^10"B"  and ""_5^11"B".

#### Solution

Mass of boron isotope ""_5^10"B", m1 = 10.01294 u

Mass of boron isotope ""_5^11"B", m2 = 11.00931 u

Abundance of ""_5^10"B", η1 = x%

Abundance of ""_5^11"B", η2 = (100 − x)%

Atomic mass of boron, m = 10.811 u

The atomic mass of boron atom is given as:

"m" = ("m"_1η_1 + "m"_2η_2)/(η_1 + η_2)

10.811 = (10.01294 xx "x" + 11.00931 xx (100 -"x"))/("x" + 100 - "x")

1081.11 = 10.01294 x + 1100.931 − 11.00931 x

∴ x = (19.821)/0.99637

= 19.89%

And 100 − x = 80.11%

Hence, the abundance of ""_5^10"B" is 19.89% and that of ""_5^11"B" is 80.11%.

Concept: Atomic Masses and Composition of Nucleus
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.1 (b) | Page 462
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 1.2 | Page 462

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