E = h f = h c/lambda h = 6.6 times 10^-34 J s e = 3 times 10^8 m/s lambda = h/p
ID: 1070488 • Letter: E
Question
E = h f = h c/lambda h = 6.6 times 10^-34 J s e = 3 times 10^8 m/s lambda = h/p delta times delta p greaterthanequalto h/2 pi E_n n^2 h^2/8 m L^2 E_n = -13.6 eV Z^2/n^2 = -2.18 times 10^-18 J Z^2/n^2 For a helium atom, an electron is in the third excited state (above the ground state). What are the possible values for the orbital quantum numal For the largest value, what are the possible value for the magnetic quantum number m_t? What are the possible value for the spin quantum number m_s? For the orbital corresponding to the value of n and the largest value of t, how many electrons can the orbital hold? What is the ionization energy of the electron?Explanation / Answer
He(z=2) electronic configuration is 1s2
(a) Third excited state means 3rd shell from 1 st shell, so it is 4th shell.
Orbital quatum number are, 0 to n-1 = 0 to 3 = 0, 1 , 2, 3
(b) For largest 'l' value i.e. 3,
the possible 'ml' values are + l to - l = + 3 to - 3 = + 3, + 2, + 1, 0 , - 1 , - 2, - 3
(c) The possible values of spin quantum number are + 1/2 , - 1/2
(d) n = 4, l = 3 has ml = 7 values that means there 7 orbitals.
And we know according to Pauli's exclusive principle, each orbital can accommodate a maximum of two elecrons only.
So, total number of electrons = 7(2) = 14 electrons.
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