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What are the respective number of `alpha` and `beta`-particles emitted in the following radioactive decay?A. (a) 6 and 8B. (b) 6 and 6C. (c) 8 and 8D. (d) 8 and 6 |
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Answer» Correct Answer - D Let `n-alpha` particles and `m-beta` particles are emitted. Then, `90-2n+m=80` …(i) `200-4n=168` …(ii) Solving Eqs. (i) and (ii), we get `n=8` and `m=6` |
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