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What is the maximum number of emission lines when the excited ...

Mar 17, 2019 ... Electron doesn't get stuck forever on any of the levels with n>1. ... excited atom, but an ensemble of many and many excited hydrogen atoms. ... a cascade of quantized steps of energy loss, say, 6→5→1 or 6→4→2→1. ... To determine N, you need to sum the states, as Soumik Das rightfully commented:.

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Solution for Student Worksheet: Energy Levels in the Atom

May 7, 2015 ... The equation for determining the energy of any state (the nth) is as follows: ... of the photon released when an electron around a hydrogen atom ... in the energies of the fourth (n=4) and second (n=2) levels ... Use the relationship between a photon's energy and its wavelength to calculate the wavelength of ...

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Photochemical reaction | chemical reaction | Britannica

Ozone protects Earth's surface from intense, deep ultraviolet (UV) irradiation, ... oxygen (O2) into individual oxygen atoms, followed by subsequent reaction of ... not the energy content (i.e., the wavelength, colour, or frequency) of the radiation. ... number of absorbed photons per second, and not their energy, that determine  ... Qalaxia Master Bot
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Bohr's Model of the Hydrogen Atom – University Physics Volume 3

Explain the difference between the absorption spectrum and the emission spectrum of ... The emission spectrum of atomic hydrogen: The spectral positions of emission lines ... An empirical formula to describe the positions (wavelengths) \ lambda ... is the energy of either an emitted or an absorbed photon with frequency f.

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Given that

Energy levels(quantum number) n = 1,2,3,4

Step 1: Get an expression for wavelength of electron using the Rydberg formula

Rydberg formula

$h\upsilon_{12}=\frac{me^4}{8\epsilon_0^2h^2}[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

where, Rydberg constant R = \frac{m e^4}{8 \epsilon_0^2 h^2} = 13.6 eV = 21.76 *10^{-19} J

$h\upsilon_{12}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$h\upsilon_{12}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{hc}{\lambda_{12}}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$                                   \because \upsilon = \frac{c}{\lambda}

$\lambda_{12}=\frac{hc}{R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]}$

Plugging the known values in the above equation

Higher energy level n_2=4, Lower energy level n_1=1, planck constan h = 6.64 * 10^{-34} , speed of light c = 3* 10^{8}

$\lambda_{12} = \frac{6.64 * 10^{-34} * 3 * 10^{8}}{21.76 * 10^{-19} [\frac{1}{1^2} - \frac{1}{4^2}]}$

$\lambda_{12} = \frac{1.992*10^{-34}}{21.76 * 10^{-19} [\frac{16 - 1}{16}]}$

\lambda_{12} = 9.1544 * 10^{-8} * \frac{16}{15}

\lambda_{12} = 9.76 * 10^{-8}

Hence, wavelength of the photon \lambda_{12}=9.76*10^{-8} m

Step 2: Calculating the frequency of the photon

Frequency \upsilon = \frac{c}{\lambda_{12}}

\upsilon = \frac{3* 10^8}{9.76 * 10^{-8}}

\upsilon = 3.1*10^{15} Hz

Hence, frequency of the photon \upsilon = 3.1*10^{15} Hz