Question

$E_n=-313.6/n^{2}kcal/mole$. If the value of $E=-34.84kcal/mole$, to which value does ' $n$ ' correspond?

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (B)

$E_n=\frac{-313.6}{n^2}$
$E=-34.84$
$\therefore -34.84=\frac{-313.6}{n^2}$
$n^2=\frac{-313.6}{-34.84}=9$
$n=\sqrt{9}=3$
$n=3$