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Which series of hydrogen spectrum will have shortest wavelength ...

Transition between lyman and balmer will have shortest wavelength Dont learn ... Apurva N. Saraogi, Technical Executive at CSEA, IIT Guwahati (2017-present) ... that wavelength of lyman is always shorter than balmer than paschen followed by ... In the spectral series of hydrogen, as the wavelength decreases the lines ...

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Hydrogen spectral series - Wikipedia

The Paschen lines all lie in the infrared band. ... overlaps with the next (Brackett) series, i.e. the shortest line in the Brackett series has a wavelength that falls among the Paschen series.

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Spectral line series | physics | Britannica

Feb 25, 2021 ... Spectral line series, any of the related sequences of wavelengths ... in spacing, coming closer together toward the shortest wavelength, called the series limit. ... The Lyman series lies in the ultraviolet, whereas the Paschen, ...

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Spectral Series- Explained along with Hydrogen spectrum, Rydberg ...

Visit BYJU'S to know about Rydberg formula, Hydrogen spectral series like, Lyman, ... Physics Important Questions ... 𝜆 is the wavelength; R is the Rydberg constant has the value 1.09737✕107 m-1; Z is the atomic number ... All the wavelength of Paschen series falls in the Infrared region of the electromagnetic spectrum.

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Rydberg formula $As per the third postulate of Bohr's model, if an atom moves from a higher energy state with quantum number \left(n_2\right) to a lower energy state with quantum number\left(n_1\right), the difference in energy is taken away by a photon of frequency \upsilon_{hl} . The frequency of any line in a series is a two-term difference. [math]h\upsilon_{hl}=\frac{me^4}{8\epsilon_0^2h^2}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ where, Rydberg constant R = \frac{m e^4}{8 \epsilon_0^2 h^2} = 13.6 eV = 21.76 *10^{-19} J

A line spectrum of different series is emitted by the atomic hydrogen.

Paschen series: $\upsilon=R\left[\frac{1}{3^2}-\frac{1}{n_2^2}\right]$;   n_2 = 4, 5, 6............

Step 1: Set up an equation for shortest wavelengths of Paschen series

$h\upsilon_{hl}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{hc}{\lambda_{hl}}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$                                      \because \upsilon = \frac{c}{\lambda}

$\lambda_{hl}=\frac{hc}{R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]}$

Step 2: plug in the known values in the above equation

Lower quantum number n_1 = 3 , Higher quantum number \ n_2=\infty, Planck constant h = 6.64 * 10^{-34} , speed of light c = 3* 10^{8}

$\lambda_{hl}=\frac{6.64*10^{-34}*3*10^8}{21.76*10^{-19}[\frac{1}{3^2}-\frac{1}{\infty^2}]}$

\lambda _{hl} = \frac{1.992 * 10^{-25}}{2.42*10^{-19}}

\lambda _{hl} = 8.23 * 10^{-7} m

Hence, shortest wavelength in Paschen series   \lambda _{hl} = 8.23 * 10^{-7} m