- A$34$
- ✓$54$
- C$100$
- D$50$
If $\frac{{{K_2}}}{{{K_1}}} = 2$
$\log 2 = \frac{{{E_a}}}{{2.303 \times 8.314}}\left[ {\frac{1}{{300}} - \frac{1}{{310}}} \right]$
${E_a} = .3010 \times 2.303 \times 8.314\left( {\frac{{300 \times 310}}{{10}}} \right)$
$ = 53598.59\;J\,mo{l^{ - 1}}$
$ = 54\;kJ$.
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| Energy level | $K$ | $L$ | $M$ | $N$ |
| $n=1$ | $n=2$ | $n=3$ | $n=4.....n=\infty$ | |
| Energy | $- 864 \,a.u.$ | Zero |
the excitation energy needed to raise the electron from $M$ level to $n = \infty$ would be :




$\text { A } \quad\quad\quad\quad\quad \text { B } \quad\quad\quad\text { C } \quad\quad\quad\quad\text { D }$
$1 \times 10^{-4} \quad 2 \times 10^{-4} \quad 0.1 \times 10^{-4} \quad 0.2 \times 10^{-4}$
(Where $E$ is the electromotive force)
Which of the above half cells would be preferred to be used as reference electrode?