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Question

The de Broglie wavelength of electron in the second orbit of Li2+ ion will be equal to the de Broglie wavelength of electron in

A
n = 3 of H atom
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B
n = 4 of C5+ ion
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C
n = 6 of Be3+ ion
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D
n = 3 of He+ ion
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Solution

The correct option is B n = 4 of C5+ ion
de-Broglie wavelength is inversely proportional to velocity. When two electrons have the same velocity, they have same de-Broglie wavelength.
Velocity is directly proportional to the ratio Zn. Here, Z is the atomic number and n is the number of the orbit in which electron is present.
For Li2+ ion, Zn=32
For C5+ ion, Zn=64=32​.

Since, Zn.​ is the same for both ions, they will have the same velocity and hence same De-Broglie wavelength.

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