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Beta Decay

A positron is...

Question

A positron is emitted from $_{11} N a^{24} .$ The ratio of the atomic mass and atomic number of the resulting nuclide is

(IIT-JEE, 2007)

22/10

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22/11

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23/10

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23/12

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Solution

The correct option is C
23/10

$_{11} N a^{23} ⟶_{10} N a^{23} +_{- 1} e^{0}$
Ratio of $\frac{A t o m i c m a s s}{A t o m i c n u m b e r}$ $= \frac{23}{10}$

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A positron is emitted from $_{11} N a^{24} .$ The ratio of the atomic mass and atomic number of the resulting nuclide is

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238_{92}^{238} U

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