Enter your answer in the provided box in scientific notation. A hydrogen-like io

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 × 1018J) 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

Explanation / Answer

Ans. Electrons are first removed from outermost shell then gradually from the inner shells. So, the last wo electrons would always be in the first shell (n = 1 ; as 1s2 electrons) because 1s in the innermost shell.

Therefore, n = 1

Atoms number remains constant at 80.

# Energy of electron for n = 1 , Z= 80

En = - (2.18 x 10-18 J) x Z2 x (1 / n2)

            Or, En = - (2.18 x 10-18 J) x (80)2 x (1 / 12)

            Hence, En = - 1.3952 x 10-14 J.

# For ionization of Hg79+ into Hg80+, the electron in n = 1 shell must be provided (say, from outside) an energy equal to its own energy to escape from the atom.

So,

The required ionization energy = + 1.3952 x 10-14 J.

Note: the +ve sign indicates that energy must be provided for ionization to occurs.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.