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Question: de-Broglie wavelength of electron in second orbit of Li2+ ion will be equal to de-Broglie of wavelen...

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

A

n = 3 of H-atom

B

n = 4 of C5+ ion

C

n = 6 of Be3+ ion

D

n = 3 of He+ ion

Answer

n = 4 of C5+ ion

Explanation

Solution

}{\frac{2}{3} = \frac{4}{6}\left( n = \frac{2}{3} = \frac{4}{6} \right)\left( n = 4ofC^{5 +}ion \right)}$$