A positron is emitted from 2311Na- The ratio of the atomic mass and atomic number of the resulting nuclide is-

The required nuclear reaction is ^11Na^23→ _+1e^0 + _10Na^23 ...

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Byju's Answer

Standard XII

Chemistry

Beta Decay

A positron is...

Question

A positron is emitted from $_{11}^{23} N a$ . The ratio of the atomic mass and atomic number of the resulting nuclide is:

\frac{22}{10}

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\frac{22}{11}

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\frac{23}{10}

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\frac{23}{12}

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Solution

The correct option is C $\frac{23}{10}$
The required nuclear reaction is
$^{11} N a^{23} \to_{+ 1} e^{0} +_{10} N a^{23}$

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A positron is emitted from 2311Na. The ratio of the atomic mass and atomic number of the resulting nuclide is:

The correct option is C 2310The required nuclear reaction is 11Na23→+1e0+10Na23

A positron is emitted from $_{11}^{23} N a$ . The ratio of the atomic mass and atomic number of the resulting nuclide is:

The correct option is C $\frac{23}{10}$
The required nuclear reaction is
$^{11} N a^{23} \to_{+ 1} e^{0} +_{10} N a^{23}$