(solved)Question 2.34 of NCERT Class XI Chemistry Chapter 2

Calculate the energy required for the process He+ (g) → He2+ (g) + e- The ionization energy for the H atom in the ground state is 2.18 × 10-18 J atom-1
(Rev. 31-Oct-2024)

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Parveen,

Question 2.34
NCERT Class XI Chemistry

Calculate the energy required for the process He+ (g) → He2+ (g) + e- The ionization energy for the H atom in the ground state is 2.18 × 10-18 J atom-1

Solution in Detail
(video solution below this)

by Bohr's theory, IE is the energy
required to remove an electron from
the n-th shell of a H-like atom of
atomic number Z:

En=+RH(Z2n2)\displaystyle E_n = +R_H\bigg(\frac{Z^2}{n^2}\bigg) J/atom

∴ IE of hydrogen for n = 1 and Z = 1 is

E1=RH(1212)=RH (J/atom)\displaystyle E_1 = R_H \bigg(\frac{1^2}{1^2}\bigg) = R_H \text{ (J/atom)}

but given E1=2.18×1018 J/atom\displaystyle E_1 = 2.18 \times 10^{-18}\text{ J/atom}

RH=2.18×1018 . . . (1)\displaystyle \therefore R_H = 2.18 \times 10^{-18}\text{ . . . (1)}

He+ is hydrogen-like with 1 electron

hence, IE for He+ (Z = 2, n = 1)

 IE =+RH(2212)=RH×4\displaystyle \therefore \text{ IE } = +R_H\bigg(\frac{2^2}{1^2}\bigg) = R_H \times 4

=2.18×1018×4[RH by (1)]\displaystyle = 2.18\times10^{-18} \times 4 \:\bigg[\text{R}_H \text{ by (1)}\bigg]

=8.72×1018 J/atom Ans\displaystyle = 8.72 \times 10^{-18} \text{ J/atom }\underline{Ans}

Video Explanation

Please watch this youtube video for a quick explanation of the solution:

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