Question

What is the radius of a hydrogen atom whose electron moves at$7.3\times {10}^5\mathrm{m}/\mathrm{s}$ ?

Short answer

The radius of the hydrogen atom $=0.476\mathrm{nm}$

Step-by-step solution

given information

An electron, in a certain state n, moves with a speed equals to $7.3\times {10}^5\mathrm{m}/\mathrm{s}$around the hydrogen's nucleus

simplification

Our aim is to find the radius of the hydrogen atom. An electron, moving in a circular orbit $r$around a certain nucleus, makes a standing waves where

$2\pi r_n=n\lambda$

Every particle, which has a certain momentum, has an associated wavelength where

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

By substituting (2) into (1), we get

$2\pi r_n=n\frac{h}{m_c{v}_n}\Rightarrow r_n=\frac{nh}{2\pi m_c{v}_n}$

From Fig.(38.27) in the book, we know that.

role="math" localid="1650742362891" $r_n=n^2{a}_B\text{(where}{a}_D\text{is the bohr radius)}$

By equating (3) and (4), we get

role="math" localid="1650742483141" $$\begin{gathered} n^2{a}_B=\frac{yh}{2\pi m_e{v}_n}\to n=\frac{h}{2\pi m_e{v}_n{a}_{Hs}} \\ \simeq 3 \end{gathered}$$

Hence the radius of the hydrogen atom, in which an electron moves with a speed equals to $\mathbf{7}.\mathbf{3}\times \mathbf{105}\mathrm{m}/\mathrm{s}$in the third state, is

role="math" localid="1650742623215" $$\begin{gathered} r_3=3^2\times 5.29\times {10}^{-2}\mathrm{nm} \\ =0.476\mathrm{nm} \end{gathered}$$