Question

If the ionization energy of hydrogen is 313.8 kcal per
mole, then the energy of the electron in 2nd excited
state will be :-
(1)-113.2 kcal/mole
(3)-313.8 kcal/mole
(2) -78.45 kcal/mole
(4) -35 kcal/mole
​

Answer 1

Answer:

option (4)

Explanation:

the ionization energy of hydrogen is 313.8 kcal per  mole.

We have to find the energy of electron in 2nd excited state,

second excitation state-----> n = 3

Energy of the electron is given by,

$E = -313.8 \ \frac{Z^{2} }{n^{2} } kcal/mole\\\\E = -313.8 \ \frac{1^{2} }{3^{2} } \\\\E = \frac{-313.8}{9} \\\\ E=-34.9 \ kcal/mole\\\\E=-35 \ kcal/mole$