Calculate the de Broglie wavelength associated with the electron in the 2nd excited state of hydrogen atom. The ground state energy of the hydrogen atom is 13.6 eV.
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A: solution is given by
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A: The energy in a hydrogen atom is given by: En=-21.8×10-19Jn2Z2 Z=1 En=-21.8×10-19n2
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A: we have n2 = 2 λ=434nm = 434×10-9m RH = 1.09737316×107 m-1
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A: Here, 1λ=R1n12-1n22
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A: n1=4 n2=6
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A: Given: The quantum number of the orbit is 3.
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A: Given : Bhor model. Ground state n =1, n = 1 Difference in radius, rnx - r1 = 5.247 x 10-9
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A: Given, Potential energy of electron proton in an atom, U = 15.4 eV 1 eV = 1.6×10-19 C
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Q: Which is not a solution of the Schrodinger equation? Question 13 options: W=eikx リ=e-kx W=sin kx…
A: As this is multiple question as per the guidelines we are going to answer first question only.…
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