Question

Energy required to ionize $2$ mole of gaseous $H{e^ + }$ ion present in its ground state is
A. $54.4eV$
B. $108.8{N_A}eV$
C. $54.4{N_A}eV$
D. $108.8eV$

Answer 1

For dealing these type questions we have to know the particular formula for which the problem can be evaluated and here it is given to find the ionization energy. Energy required to remove an electron from the ground state of an atom which is the same as hydrogen is given by ionization energy. Here , we have to find the particular energy required to ionize two moles of gaseous $H{e^ + }$ ion present in the ground state.
Formula used: $+ 13.6\dfrac{{{Z^2}}}{{{n^2}}}$

Complete step by step solution:
Energy required to ionize $2$ mole of gaseous $H{e^ + }$ ion present in its ground state is given by,
E = $$$$ + 13.6\dfrac{{{Z^2}}}{{{n^2}}}
Where $E$ is the ionization energy ,
$Z$ is the atomic number ,
$n$ is the number of electrons.

Here $Z$ which is the atomic number of helium atom = $2$
$n$ is the number of electrons = $1$ ($H{e^ + }$ is in ionic state with losing 1 electron since its atomic number is$2$ ).
Substitute the values in the above equation ,
$E = 13.6 \times 4$
$E = 54.4eV$ atom

Number of moles = Number of atom divided by ${N_A}$
Number of moles is $2$ which is given in the question.
Number of atom $= 2 \times {N_A}$
Therefore , ${E_{total}} = 54.4 \times 2{N_A}eV$
$= 108.8{N_A}eV$

Hence option B is the correct answer.

Note: Ionizing energy is used for withdrawing an electron from a gaseous atom or ion. In the periodic table, we can see ionization energy increases from left to right and ionization energy decreases from top to bottom. When ionization energy is supplied to an atom of element , a greater nuclear charge pulls electrons close to the nucleus, which thus decrease the atomic radius.