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2 years ago

What is the condensed electron configuration for nitrogen

Chemistry

2 answers:

Alenkinab [10]2 years ago

6 0

Answer:

In writing the electron configuration for nitrogen the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for N goes in the 2s orbital. The remaining three electrons will go in the 2p orbital. Therefore the N electron configuration will be 1s22s22p3.

Explanation:

hope you get it right:)

Debora [2.8K]2 years ago

3 0

Answer:

the electron configuration for N will be 1s^2 2s^2 2p^3

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Answer:

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Explanation:

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The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

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Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

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$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

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$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

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$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

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