Question

Enter your answer in the provided box in scientific notation. A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by

En = −(2.18 × 10−18J) Z2 (1/n2)

where n is the principal quantum number and Z is the atomic number of the element.

Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step:

Hg79+(g) → Hg80+(g)+ e−

Answer 1

For Hg⁸⁰⁺, Z = 80, and you should assume ground state (n=1) for the outermost (only) electron.

En = −(2.18 × 10⁻¹⁸ J) Z² (1/n²)

E = -[-2.18 x 10^-18 J x (80)² x 1/1²]= 1.3952 x 10^-14 J

E = 1.3952 x 10^-17 kJ, we know 1 kJ = 1000 J so 1 J = 10^-3 kJ

but 1 mol = 6.02214129×10²³ = Avagadro's number

so, E = 6.02214129×10²³ x 1.3952 x 10^-17 kJ/mol = 8402091.52781 kJ/mol

E = 8.40 x 10⁶ kJ/mol

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