An element, $^{'} E^{'}$ has the following isotopic composition: $^{200} A = 90 %;^{199} A = 8 %;^{202} A = 2 %$ . The average atomic mass of the naturally occurring element is:

201 a m u

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202 a m u

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198 a m u

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200 a m u

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Solution

The correct option is D $200 a m u$
Average atomic mass $= \frac{\sum A t o m i c m a s s \times % a b u n d a n c e}{100}$

$= \frac{(200 \times 90 + 8 \times 199 + 2 \times 202)}{100} = 199.96 ≃ 200 a m u$

Hence, option D is correct.

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An element, ′E′ has the following isotopic composition: 200A=90%;199A=8%;202A=2%. The average atomic mass of the naturally occurring element is:

The correct option is D 200amuAverage atomic mass =∑Atomicmass×%abundance100 =(200×90+8×199+2×202)100=199.96≃200 amuHence, option D is correct.

An element, $^{'} E^{'}$ has the following isotopic composition: $^{200} A = 90 %;^{199} A = 8 %;^{202} A = 2 %$ . The average atomic mass of the naturally occurring element is:

The correct option is D $200 a m u$
Average atomic mass $= \frac{\sum A t o m i c m a s s \times % a b u n d a n c e}{100}$

$= \frac{(200 \times 90 + 8 \times 199 + 2 \times 202)}{100} = 199.96 ≃ 200 a m u$

Hence, option D is correct.