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The ionization enthalpy of hydrogen atom is $\dpi{100} 1.312 \times 10 ^{6}J mol ^{-1}$ The energy required to excite the electron in the atom from$\dpi{100} n=1\: to\: n=2$

• 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)

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

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N

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

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