Question

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 from $n=1$ to $n=2$ is

• A. $9.84 \times 10^{5} J mol ^{-1}$
• B. $8.51 \times 10^{5} J mol ^{-1}$
• C. $6.56 \times 10^{5} J mol ^{-1}$
• D. $7.56 \times 10^{5} J mol ^{-1}$

Answer 1

Answer: (A)

The ionisation of $H$ - atom is the energy absorbed when the electron in an atom gets excited from first shell $\left(E_{1}\right)$ to infinity $\left(\right.$ i.e. $\left., E_{\infty}\right)$.

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

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

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

$E_{2}=-\frac{1.312 \times 10^{6}}{(2)^{2}}=-\frac{1.312 \times 10^{6}}{4} J mol ^{-1}$

Energy of electron in second orbit $(n=2)$

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

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

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

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