Question

Supposing the ionisation energy of the Hydrogen atom is 640eV. Point out the main shell having energy equal to 40eV-
(1) n=2
(2) n=3
(3) n=4
(4) n=5

Answer 1

Answer: -

If the ionisation energy of the Hydrogen atom is 640eV. the n = 4 main shell having energy equal to 40eV.

Explanation: -

For hydrogen atom, Z = 1.

Energy for nth main shell of hydrogen atom = Ionization Energy of Hydrogen atom / n²

40 = 640/ n²

n² = $\frac{640}{40}$

n = 4

Thus, if the ionisation energy of the Hydrogen atom is 640eV. the n = 4 main shell having energy equal to 40eV.

Answer 2

If the ionisation energy of the Hydrogen atom is 640eV. The n = 4 main shell having energy equal to 40eV.

For hydrogen atom, Z = 1.

Energy for nth main shell of hydrogen atom = (Ionization Energy of Hydrogen atom / n²)

40 = (640/ n²)

n² = (640/40)

n = 4

Thus, if the ionisation energy of the hydrogen atom is 640eV. The n = 4 main shell having energy equal to 40eV.

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