What is the position of an element in the periodic table if its electron configuration is 1s2 2s2 2p6 3s2 3p5
2 answers:
Answer : The position of an element in the periodic table is, Group 17 and Period 3
Explanation :
The given electronic configuration is,
Total number of electrons = 17
Number of valence electrons = 2 + 5 = 7
In the given configuration, 3 is the highest principle quantum number. Therefore, it belongs to the 3rd period in the periodic table.
If the elements belongs to the p-block then we have to determine the group number by,
Group number = Number of valence electrons + 10
Group number = 7 + 10 = 17
Hence, the position of an element in the periodic table is, Group 17 and Period 3
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Answer:
The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.
Explanation:
In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:
, where:
[tex]m_{e}[tex] = electron mass
e = electron charge
[tex]\epsilon_{0}[tex] = vacuum permittivity
[tex]\hbar[tex] = Planck's constant over 2pi
n = quantum number
[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV
Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:
[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]
Now, knowing the energy of the photon, we can calculate its wavelength using the equation:
[tex]E = \frac{hc}{\lambda}[tex], where:
E = Photon's energy
h = Planck's constant
c = speed of light in vacuum
[tex]\lambda[tex] = wavelength
Solving for [tex]\lambda[tex] and substituting the required values:
[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).