Concept explainers
(a)
Interpretation:
The transition of an electron form ground state He+ to a state with quantum numbers, n=4, l=1, ml=+1 has to be described using term symbols.
Concept introduction:
A term symbol describes the orbital, spin and total angular momenta of an electronic state. The symbol L represents the vector sum of orbital
2S+1LJ
(a)
Answer to Problem 8.1IA
The transition of an electron from ground state He+ to a state with quantum numbers, n=4, l=1, ml=+1 is represented as 2S1/2→2P1/2.
Explanation of Solution
The term symbol for the He+ is written as shown below.
2S+1LJ..............(1)
Where,
L is the quantum number indicating sum of orbital angular momentum
S is the spin angular momentum
J is the total angular momentum.
The value of L is equal to the value of l.
The value of l in the excited state is 1.
Therefore the value of L is 1 which corresponds to the term P.
The value of S for He+ ion in the excited state is 12
Therefore, the value of 2S+1 is calculated below.
2S+1=2×12+1=2
As the orbital is less than half filled in excited state He+, the total angular momentum (J) for He+ is given by the formula below.
J=|L−S|........................(2)
Substitute the value of L and S in equation (1).
J=1−12=12
Substitute the value of S, J and the term for L=1 in equation (1).
Therefore the term symbol for the excited state of He+ is shown below.
2P1/2
The ground state configuration of He+ is 1s1.
The value of L is equal to the value of l.
The value of l for the s orbital is 0.
Therefore the value of L is 0 which corresponds to the term S.
The value of S for He+ atom in the excited state is 12
Therefore, the value of 2S+1 is calculated below.
2S+1=2×12+1=2
The total angular momentum (J) for He+ in the ground state is given by the formula below.
J=|L−S|........................(2)
Substitute the value of L and S in equation (1).
J=|0−12|=12
Substitute the value of S, J and the term for L=0 in equation (1).
Therefore the term symbol for the excited state of He+ is shown below.
2S1/2
Therefore, the transition occurs from 2S1/2→2P1/2.
(b)
Interpretation:
The wavelength, frequency and wavenumber of the transition has to be calculated.
Concept introduction:
The wavelength, frequency and wavenumber of transition are all related to the energy of the transition. The wavelength is inversely proportional to the energy of transition. The frequency and wavenumber are directly proportional to energy of transition.
(b)
Answer to Problem 8.1IA
The frequency of the transition is 1.23×1016 s−1_. The wavelength of the transition is 2.43×10−8 m_. The wavenumber of the transition is 0.411×108 m−1_.
Explanation of Solution
The energy of transition of the hydrogen like atom is given below.
En→n'=13.6Z2(1n2−1(n')2) eV..............(3)
Where,
n and n' is the number of the state.
Z is the
The atomic number of He+ (Z) is 2.
The transition takes place from n=1 to n'=4.
Substitute the value of n, and n' in equation (3).
En→n'=13.6×(2)2(11−1(4)2) eV=54.4×1516 eV=51 eV×1.6×10−19 J1eV=8.16×10-18 J
The formula to calculate frequency (v) is given below.
E=hv...............(4)
Where,
h is Planck’s constant.
The value of Planck’s constant (h) is 6.626×10−34 J s.
Substitute the value of E in equation (4).
8.16×10−18 J=6.626×10−34 J s×vv=8.16×10−18 J6.626×10−34 J sv=1.23×1016 s−1_
Therefore, frequency of the transition is 1.23×1016 s−1_.
The wavelength (λ ) is calculated by the formula below.
λ=cv..............(5)
Where,
c is the speed of light.
The speed of light is 3×108 m s−1.
Substitute the value of v and c in equation (5).
λ=3×108 m s−11.23×1016 s−1=2.43×10−8 m_
Therefore, wavelength of the transition is 2.43×10−8 m_.
The relation wavelength (λ) and wavenumber (˜ν) is given below.
1λ=˜ν..................(6)
Substitute the value of λ in equation (2).
˜ν=12.43×10−8 m=0.411×108 m−1_
Therefore, the wavenumber is 0.411×108 m−1_.
(c)
Interpretation:
The change in mean radius of electron during the transition has to be calculated.
Concept introduction:
The total distance from the nucleus of an atom to its outermost shell, where probability of finding the electron is maximum is termed as radius of the atom. The most probable distance of the electron form nucleus in the ground state of hydrogen atom is termed as Bohr’s radius.
(c)
Answer to Problem 8.1IA
The change in mean radius during transition is 10.75a0.
Explanation of Solution
The mean radius of the electron is given by the formula shown below.
rn,l,ml=n2a0Z{1+12[1−l(l+1)n2]}..................(7)
Where,
Z is the atomic number.
a0 is the Bohr’s radius.
The atomic number of He+ (Z) is 2.
Substitute the value of Z, n=4 and l=1 in equation (7).
rn,l,ml=42a02{1+12[1−1(1+1)42]}=8a0{1+12[1−18]}=8a0{2316}=23a02
The ground state configuration of He+ is 1s1.
The value of l for the s orbital is 0.
The value of n is 1.
Substitute the value of Z, n=1 and l=0 in equation (7).
rn,l,ml=12a02{1+12[1−0(0+1)n2]}=1a02{1+12}=3a04
Therefore, the change in radius during transition is calculated below.
Change in radius=23a02−3a04=43a04=10.75a0
Therefore, the change in mean radius during transition is 10.75a0.
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