Question

The de-broglie wavelength of electron revolving in the $n_{th}$ orbit of ${He}^{+}$ ion is 6.64$\mathring{A}$. The value of 'n' is -

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

Velocity of the electron in $H{e}^{+}$ ion in the n${}^{th}=2.18\times {10}^6\times \frac{Z}{n}$ m/s=$\frac{4.36\times {10}^6}{n}$

As $\lambda =\frac{h}{mv}$

$⟹m\left(\frac{4.36\times {10}^6}{n}\right)\times 6.64\times {10}^{-10}=6.626\times {10}^{-34}⟹9.1\times {10}^{-31}\times \frac{4.36\times {10}^{-4}}{n}={10}^{-34}⟹n=0.91\times 4.36=4$

The orbit is 4${}^{th}$ orbit.