Question

De-Broglie wavelength of electron in second orbit of $L{i}^{2+}$ ion will be equal to de-Broglie of wavelength of electron in:

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

Answer 1

Answer: (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.

Hence, the correct option is $B$