Question

The ionization enthalpy of hydrogen atom is 1.312×10⁶ j mol^-1. The energy required to excite the electron in the atom from n=1 to n=2 is:

• A. 8.51×10⁵ J mol^-1
• B. 6.56×10⁵ J mol^-1
• C. 7.56×10⁵ J mol^-1
• D. 9.84×10⁵ J mol^-1

Answer 1

The correct option is D

9.84×10⁵ J mol^-1

($△$E),The energy required to excite an electron in an atom of hydrogen fron n=1 to n=2 is

(difference in energy $E_2$ and $E_1$)

Values of $E_2$ and $E_1$ are,

$E_2$=$\frac{-1.312\times {10}^6\times (1{)}^2}{(2{)}^2}$=-3.28$\times {10}^5$ J $mo{l}^{-1}$

$△$E is given by the relation,

$E_1$=$-1.312\times {10}^6$ J $mo{l}^{-1}$

$△$E= $E_2$-$E_1$

=[$-3.28\times {10}^5$ J $mo{l}^{-1}$]-$[-1.312\times {10}^6$ J $mo{l}^{-1}$]

=($-3.28\times {10}^5$+$1.312\times {10}^6$) $Jmo{l}^{-1}$

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

Thus the correct answer is (D).