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Standard XII

Chemistry

Scales of Electronegativity

If the ioniza...

Question

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

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Solution

Step 1:
$2^{n d}$ excited state means n = 3
Energy varies inversely with square of number of excited state, i.e.,
$E_{n} = - \frac{E_{1}}{n^{2}}$
Step 2:
$E_{2} = - \frac{313.8}{9} = - 34.87 K c a l {m o l}^{- 1}$
Hence, the energy of second excited state will be -34.87 Kcal/mol

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

Step 1: 2nd excited state means n = 3 Energy varies inversely with square of number of excited state, i.e., En=−E1n2 Step 2: E2=−313.89=−34.87Kcalmol−1 Hence, the energy of second excited state will be -34.87 Kcal/mol

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

Step 1:
$2^{n d}$ excited state means n = 3
Energy varies inversely with square of number of excited state, i.e.,
$E_{n} = - \frac{E_{1}}{n^{2}}$
Step 2:
$E_{2} = - \frac{313.8}{9} = - 34.87 K c a l {m o l}^{- 1}$
Hence, the energy of second excited state will be -34.87 Kcal/mol