If 917 $\mathop A\limits^o$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be ___________ $\mathop A\limits^o$.

<p>The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by </p> <p>$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $</p> <p>where ($n_1$) and ($n_2$) are the initial and final energy levels, respectively. For the Lyman series, the electron transitions to the ground state (($n_1$ = 1)), and for the Balmer series, the electron transitions to the first excited state (($n_1$ = 2)).</p> <p>From the given information, we have the lowest wavelength in the Lyman series, ($\lambda_1 = 917 \, \text{Å}$). Therefore, the energy difference ($\Delta E$) for the Lyman series is </p> <p>$ \Delta E = \frac{hc}{\lambda_1} $</p> <p>where (h) is Planck&#39;s constant and (c) is the speed of light.</p> <p>Similarly, for the Balmer series, the energy difference ($\Delta E$) is </p> <p>$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = -13.6 \, \text{eV} \times \frac{1}{4} $</p> <p>The corresponding wavelength ($\lambda_2$) is </p> <p>$ \lambda_2 = \frac{hc}{\Delta E} $</p> <p>By comparing ($\Delta E$) for the Lyman and Balmer series, you found that </p> <p>$ \frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{1}{4} $</p> <p>Therefore, the lowest wavelength in the Balmer series is </p> <p>$ \lambda_2 = 4 \lambda_1 = 4 \times 917 \, \text{Å} = 3668 \, \text{Å} $</p>