Question

Supposing I.P. of H-atom is 270 eV. Find out the value of principal quantum number in which electron has the energy equal to -30 eV :

(1) n = 2 (3) n = 4 (3) n = 4

(2) n = 3 (4) n = 1

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Answer 1

Given:

• The ionization potential of H-atom = 270 eV
• The energy of the electron in the nth orbit of the H-atom (Eₙ) = -30 eV

To find:

The value of n.

Solution:

• Since, during ionization the transition of electron takes place from n=1 to n=∞ and the energy of the electron at n=∞ is equal to zero, the energy of electron in the ground state of hydrogen atom (E₀) = -270 eV.
• The energy of electron in the nth orbit (Eₙ) = E₀/n² ⇒ n = √(E₀/Eₙ) = √9 = 3

Answer:

The value of n = 3.