EN of element A is E_1 and IP is E_2 hence EA will be - Vidyadip

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
$\frac{\mathbb{P}+\mathrm{EA}}{2}=\mathrm{EN} ; \frac{\mathrm{E}_{2}+\mathrm{EA}}{2}=\mathrm{E}_{1}$

hence $\mathrm{EA}=2 \mathrm{E}_{1}-\mathrm{E}_{2}$

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