MCQ
$EN$ of element $A$ is $E_1$ and $IP$ is $E_2$ hence $EA$ will be
- ✓$2E_1 -E_2$
- B$E_1 -E_2$
- C$E_1 -2E_2$
- D$(E_1 + E_2)/2$
✓
Answer
Correct option: A.
$2E_1 -E_2$
a
$\frac{\mathbb{P}+\mathrm{EA}}{2}=\mathrm{EN} ; \frac{\mathrm{E}_{2}+\mathrm{EA}}{2}=\mathrm{E}_{1}$
$\frac{\mathbb{P}+\mathrm{EA}}{2}=\mathrm{EN} ; \frac{\mathrm{E}_{2}+\mathrm{EA}}{2}=\mathrm{E}_{1}$
hence $\mathrm{EA}=2 \mathrm{E}_{1}-\mathrm{E}_{2}$
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