Question 373: Atoms & Nuclei

Question

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

Accepted Answer

The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by
$
\Delta E = -13.6 \, 	ext{eV} 	imes \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight)
$
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)).
From the given information, we have the lowest wavelength in the Lyman series, ($\lambda_1 = 917 \, 	ext{ÃÂ}$). Therefore, the energy difference ($\Delta E$) for the Lyman series is
$
\Delta E = \frac{hc}{\lambda_1}
$
where (h) is Planck's constant and (c) is the speed of light.
Similarly, for the Balmer series, the energy difference ($\Delta E$) is
$
\Delta E = -13.6 \, 	ext{eV} 	imes \left(\frac{1}{2^2} - \frac{1}{\infty^2}ight) = -13.6 \, 	ext{eV} 	imes \frac{1}{4}
$
The corresponding wavelength ($\lambda_2$) is
$
\lambda_2 = \frac{hc}{\Delta E}
$
By comparing ($\Delta E$) for the Lyman and Balmer series, you found that
$
\frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{1}{4}
$
Therefore, the lowest wavelength in the Balmer series is
$
\lambda_2 = 4 \lambda_1 = 4 	imes 917 \, 	ext{ÃÂ} = 3668 \, 	ext{ÃÂ}
$

Atoms & Nuclei

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 $.

Choose the correct answer: