$\frac{\left(n_2-n_1\right)\left(n_2-n_1+1\right)}{2}=10$
$\frac{\left(n_2-n_1\right)\left(n_2-1+1\right)}{2}=10$
$\frac{n_2\left(n_2-1\right)}{2}=10$
$\because \quad n_2=5 \frac{\left(n_2-n_1\right)\left(n_2-n_1+1\right)}{2}=3$
$\frac{\left(5-n_1\right)\left(5-n_1+1\right)}{2}=3$
$\frac{\left(5-n_1\right)\left(6-n_1\right)}{2}=3$
$n_1=3 \quad\left(2^{n d}\right. \text { excite d state ) }$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
(Given, $\frac{ d (\ln K )}{ d \left(\frac{1}{T}\right)}=-\frac{\Delta H^{\ominus}}{ R }$, where the equilibrium constant, $K =\frac{ p _{ z }}{ p ^{\ominus}}$ and the gas constant, $R =8.314$ $\left.J K ^{-1} mol ^{-1}\right)$
($1$) The value of standard enthalpy, $\Delta H ^{\ominus}$ (in $kJ mol ^{-1}$ ) for the reaction is. . . . . . .
($2$) The value of $\Delta S^{\ominus}$ (in $J K ^{-1} mol ^{-1}$ ) for the given reaction, at $1000 K$ is. . . . . .
Give the answer or quetin ($1$) and ($2$)
