Question

The circumference of the second Bohr orbit of electron in the hydrogen atom is $600nm$. Calculate the potential difference to which the electron has to be subjected so that the electron stops. The electron had the de Broglie wavelength corresponding to the circumference.

Answer 1

Number of waves '$n$' $=\frac{Circumference}{Wavelength}$
$n\lambda =2\pi r$
$2\lambda =600$
$\lambda =300nm$
Let stopping potential is $V_0$.
$e{V}_0=\frac{1}{2}m{v}^2$ ...(i)
$\lambda =\frac{h}{mv}$
$v=\frac{h}{\lambda m}$ ...(ii)
From equations (i) and (ii),
$e{V}_0=\frac{1}{2}m{\left(\frac{h}{\lambda m}\right)}^2$
$V_0=\frac{h^2}{2m{\lambda }^{2}e}$
$=\frac{{(6.626\times {10}^{-34})}^2}{2\times (9.1\times {10}^{-31})\times {(300\times {10}^{-9})}^2\times 1.6\times {10}^{-19}}$
$=1.675\times {10}^{-5}V$.

Hence the potential difference to which the electron has to be subjected so that electron stops is $=1.675\times {10}^{-5}V$.