If $^{\prime} \lambda_{1}$ ' and ' $\lambda_{2}$ ' are the wavelengths of de-Broglie waves for electrons in…
If $^{\prime} \lambda_{1}$ ' and ' $\lambda_{2}$ ' are the wavelengths of de-Broglie waves for electrons in first and second Bohr orbits in hydrogen atom, then $\left(\frac{\lambda_{1}}{\lambda_{2}}\right)$ is equal to (Energy in $1^{\text {st }}$ Bohr
orbit $=-13.6 \mathrm{eV}$ )
$\frac{1}{5}$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{3}$
Solution
The de-Broglie wavelength of an electron in the $n$-th orbit is inversely proportional to the radius of the orbit. Since the radius for the first orbit is $R$ and for the second orbit is $4 R$, the de-Broglie wavelength in the second orbit is 4 times that in the first orbit.
Thus, the wavelength ratio is:
$\frac{\lambda_1}{\lambda_2}=4$
The energy in the $n$-th Bohr orbit is given by:
$E_n=-\frac{13.6 \mathrm{eV}}{n^2}$
.