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Question

Physics Question on de broglie hypothesis

The one electron species having ionization energy of 54.4 eVs54.4\, eVs

A

Be2+Be^{2+}

B

Be+3Be^{+3}

C

He+He^{+}

D

HH

Answer

He+He^{+}

Explanation

Solution

lonisation energy =−13.6Z2n2eV=-13.6 \frac{Z^{2}}{n^{2}} eV
=−13.6×(2)2(1)2=-13.6 \times \frac{(2)^{2}}{(1)^{2}} (For He+,Z=2He ^{+}, Z=2)

=−54.4 eV=-54.4\, eV