Hydrogen spectrum

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Question

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelenths of the Lyman series is 304 $\overset{0}{A}$ . The corresponding difference for the Paschan series in $\overset{0}{A}$ is : ___________ .

Moderate

Solution

Largest wavelength $\Rightarrow$ minimum energy and vice-versa.
Lyman
$\begin{array}{l} \frac{1}{λ_{\max}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \Rightarrow λ_{\max} = \frac{4}{3 R} \\ \frac{1}{λ_{\min}} = R (\frac{1}{1^{2}} - \frac{1}{\infty^{2}}) \Rightarrow λ_{\min} = \frac{1}{R} \\ λ_{\max} - λ_{\min} = \frac{1}{3 R} = 304 \overset{0}{A} \end{array}$
Paschen
$\begin{array}{l} \frac{1}{λ_{\max}} = R (\frac{1}{3^{2}} - \frac{1}{4^{2}}) \Rightarrow λ_{\max} = \frac{16 \times 9}{7 R} \\ \frac{1}{λ_{\min}} = R (\frac{1}{3^{2}} - \frac{1}{\infty^{2}}) \Rightarrow λ_{\min} = \frac{9}{R} \\ λ_{\max} - λ_{\min} = \frac{81}{7 R} = \frac{81}{7} \times 3 \times 304 \overset{0}{A} \\ = 10553.14 \overset{0}{A} \end{array}$

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Hydrogen spectrum

Remember concepts with our Masterclasses.

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelenths of the Lyman series is 304 A0. The corresponding difference for the Paschan series in A0 is : ___________ .

Largest wavelength ⇒ minimum energy and vice-versa.Lyman1λmax=R112-122⇒λmax=43R1λmin=R112-1∞2⇒λmin=1Rλmax-λmin=13R=304A0Paschen1λmax=R132-142⇒λmax=16×97R1λmin=R132-1∞2⇒λmin=9Rλmax-λmin=817R=817×3×304 A0=10553.14 A0

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In the line spectra of hydrogen atom, difference between the largest and the shortest wavelenths of the Lyman series is 304 $\overset{0}{A}$ . The corresponding difference for the Paschan series in $\overset{0}{A}$ is : ___________ .