Question

If the value of $E_n=-78.4$ kcal / mole, the order of the orbit in hydrogen atom is:

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

$E=-\frac{13.6}{n^2}eV$

$E_n=-78.4kcalmole$

$1eV=23.1kcalmole{}^{-1}$

$n^2=\frac{13.6\times 23.1}{78.4}$

$\therefore n=2$

Hence, option $C$ is correct.