Question

The de Broglie wavelength of electron in the second orbit of Li${}^{2+}$ ion will be equal to the de Broglie wavelength of electron in

• A. n = 3 of H atom
• B. n = 4 of C${}^{5+}$ ion
• C. n = 6 of Be${}^{3+}$ ion
• D. n = 3 of He${}^{+}$ ion

Answer 1

The correct option is B n = 4 of C${}^{5+}$ 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 $\frac{Z}{n}$. Here, Z is the atomic number and n is the number of the orbit in which electron is present.
For $L{i}^{2+}ion,\frac{Z}{n}=\frac{3}{2}$
For $C^{5+}ion,\frac{Z}{n}=\frac{6}{4}=\frac{3}{2}$​.

Since, $\frac{Z}{n}$.​ is the same for both ions, they will have the same velocity and hence same De-Broglie wavelength.