The ionization enthalpy of hydrogen atom is 1.312 \times 10 ^{6}J mol ^{-1} The energy required to excite the electron in the atom fromn=1\: to\: n=2

  • Option 1)

       9.84 \times 10^{5}J\: mol^{-1}

  • Option 2)

    8.51 \times 10^{5}J\: mol^{-1}

  • Option 3)

    6.56 \times 10^{5}J\: mol^{-1}

  • Option 4)

    7.56 \times 10^{5}J\: mol^{-1}

 

Answers (2)

As we learnt in 

Total energy of elctron in nth orbit -

E_{n}= -13.6\: \frac{z^{2}}{n^{2}}eV

Where z is atomic number

-

The ionisation of H­ atom is the energy absorbed when the electron in an atom gets excited from first shell (E1) to infinity

\left ( i.e.E_{\infty } \right )

I.E.=E_{\infty }-E_{1}

1.312 \times 10^{6}= 0-E_{1}

E_{1} = -1.312\times 10^{6}J mol^{-1}

E_{2} = \frac{-1.312\times 10^{6}}{\left ( 2 \right )^{2}}= -\frac{1.312\times 10^{6}}{4}

Energy of electron in second orbit ( n=2)

Energy required when an electron makes transition from  n =  1 to n = 2

\Delta E=E_{2} -E_{1} = -\frac{1.312\times 10^{6}}{4}-\left ( -1.312\times 10^{6} \right )

= \frac{-1.312\times 10^{6}+5.248\times 10^{6}}{4}= 0.984\times 10^{6}

\Delta E= 9.84\times 10^{5}J \: mol^{-1}


Option 1)

   9.84 \times 10^{5}J\: mol^{-1}

Correct option

Option 2)

8.51 \times 10^{5}J\: mol^{-1}

Incorrect option

Option 3)

6.56 \times 10^{5}J\: mol^{-1}

Incorrect option

Option 4)

7.56 \times 10^{5}J\: mol^{-1}

Incorrect option

N neha

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