Question

The circumference of the $2^{nd}$ Bohr orbit of a hydrogen atom is $600\text{nm}.$ The potential difference that must be applied between the plates so that the electrons have the de-Broglie wavelength corresponding to this circumference is-

• A. $3\times {10}^{-5}\text{V}$
• B. $5\times {10}^{-5}\text{V}$
• C. $\frac{5}{3}\times {10}^{-5}\text{V}$
• D. ${10}^{-5}\text{V}$

Answer 1

The correct option is C $\frac{5}{3}\times {10}^{-5}\text{V}$

From Bohr's postulates,

$L=mv{r}_n=\frac{nh}{2\pi }$

$\Rightarrow mv=\frac{nh}{2\pi r_n}$

de - Broglie wavelength of electron in hydrogen atom is,

${\lambda }_n=\frac{h}{mv}=\frac{2\pi r_n}{n}$

For the $2^{nd}$ orbit,

Given, $n=2;2\pi r_n=600\text{nm}$

$\lambda =\frac{2\pi r_n}{n}=\frac{600\times {10}^{-9}}{2}=300\times {10}^{-9}\text{m}=3000\mathring{A}......\left(1\right)$

The de-Broglie wavelength of the electron is,

$\lambda =\frac{12.27}{\sqrt{V}}\mathring{A}......\left(2\right)$

From (1) and (2) we get,

​​​​​$3000\mathring{A}=\frac{12.27}{\sqrt{V}}\mathring{A}$

$\Rightarrow (3000{)}^2=\frac{150}{V}$

$V=\frac{150}{(3000{)}^2}=\frac{15}{9}\times {10}^{-5}$

$\therefore V=\frac{5}{3}\times {10}^{-5}\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.