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

Answer 1

Answer: (D)

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

we know that energy of $n^{th}$ orbit for hydrogen atom is

$E_n=\frac{-k}{n^2}$ where k is a constant

$△E=E_2-E_1=\left(\frac{-k}{n_2^2}\right)-\left(\frac{-k}{n_1^2}\right)=k(\frac{1}{n_1^2}-\frac{1}{n_2^2})=k(\frac{1}{1^2}-\frac{1}{2^2})$

Given :- $k=1.312\times {10}^{6}Jmo{l}^{-1}$

$\therefore △E=1.312\times {10}^6(\frac{1}{1^2}-\frac{1}{2^2})=1.312\times {10}^6(\frac{1}{1}-\frac{1}{4})$

$=1.312\times {10}^6\left(\frac{4-1}{4}\right)=1.312\times {10}^6\left(\frac{3}{4}\right)=\frac{3.936}{4}\times {10}^6=0.984\times {10}^6$

$=9.84\times {10}^5$

Ans is D $9.84\times {10}^{5}Jmo{l}^{-1}$