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

The correct option is A $9.84\times {10}^{5}J{mol}^{-1}$

$Explanation:$

From $Boh{r}^{\prime }satomicmoddel$ we have:

$\Delta E=E_{high}-E_{low}$

$\Delta E=$ The energy required to excite the electron

$E_{high}=$Higher energy level

$E_{low}=$ Lower energy level

$\because$ Electron jumps from $n=1ton=2$

$\Delta E=E_2-E_1\to \left(2\right)$

For Hydrogen atom

$E_n=\frac{-E_1}{n^2}$

$E_1=1.312\times {10}^6$(Given)

$E_2=\frac{-E_1}{2^2}=\frac{-1.312\times {10}^6}{2^2}\to \left(2\right)$

$E_1=\frac{-E_1}{1^2}=\frac{-1.312\times {10}^6}{1^2}\to \left(3\right)$

Substitute equation $\left(2\right)$ and $\left(3\right)$ in equation $\left(1\right)$ we get:

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

$\Delta E=984000Jmo{l}^{-1}=9.84\times {10}^{5}Jmo{l}^{-1}$

$\Delta E=9.84\times {10}^{5}Jmo{l}^{-1}$

$Hencethecorrectanswerisanoption\left(A\right).$