Reydberg put forward a principle he applied to find wave length of a spectrum.
The wave length of H-spectrum determined is applied to find frequency and energy of the H- spectrum.
The wave number is inversely proportional to the square of the energy level differences ( ∆n²).
ṽ∝1
∆n²
1 = RH ∆n²
ṽ
ṽ = [ 1/n₁² - 1/n₁²]RH
Since; ṽ = 1/λ
1/λ = RH[ 1/n₁² - 1/n₂² ]………………. (3)
Where;
λ= wave length
n₁ = lowest energy level
n₂ = highest energy level
RH = Reydberg constant
RH is 1.09 x 10^7 m-¹
The value of n₁ and n₂ for H-spectrum can be obtained if given number of line and number of series. The value n₁ is equal number of series given. But the value n₂ is equal to the number of line plus the number of series.
Problem # 03;
Obtain the values of the wavelength and the corresponding energy, for n₁ and n₂ of the third line in Balmer’s series.
Solution;
Number of line = Third line
Number of series = Balmer’s series
n₁ = number of series = 2
n₂ = number of line + Number of series = 3 + 2
n = 5.
But;
1/λ = RH[ 1/n₁² - 1/n₂²]
=1.09 X 10^7m-¹[1/2²-1/5²]
=1.09 X 10^7m-¹[1/4-1/25]
= 1.09 X 10^7m-¹[21/100]
λ = 4.45 x 10^-7m of third line
Recall from eqn
E = hc/λ
= 6.3E-3 Js x 3.0E-8ms-¹
4.45E-7m
E=4.25 x 10^-19J
.: the energy of balmer series above = 4.25×10^-19J
Also;
f = c
λ
f = 3.0 × 10^8ms-¹
4.45 x 10^-7m
f = 6.74 x 10¹⁴s-¹
f = 6.74 x 10¹⁴Hz
.: the frequency should be 6.74 × 10¹⁴Hz
Transition Energy
Is the energy required to shift an electron from one shell to another.
The energy required to
shift electrons from one shell to another should be equal to the energy difference between the respective shells.
The energy difference between the lowest energy level E₁ and the highest energy level E₂ is obtained by the expression;-
E = E₂– E₁……………… (1)
The trend of transition of electrons is shown as;
E = E₂ – E₁
Transition takes place from E₁ to E₂ exactly.
E > E₂ – E₁
Transition take place from E₁ toward beyond E₂
E < E₂ – E₁ – Transition take place from E₁ and hang between E₁ and E₂
But if electrons gain enough energy which is equal to the ionization energy results the
electron to jump completely from ground state to infinite. These electrons cannot return back instead leave an atom ionized positively. The energy supplied to the atom results it to be excited and electron leave completely from the ground state to infinite. The energy supplied to the electrons is used as ionization energy and as kinetic energy.
The ionization energy of electrons is equal to the amount of energy associated by electron in the shell or quantum number where it belongs.
The expression below is used to find the energy associated by electrons or energy of that shell.
1 = RH
λ
Let; n₁ = n
n₂ = ∞
1 = RH[ 1 - 1 ]
λ n² ∞²
1 = RH[ 1 ] since; 1 = 0
λ n² ∞²
Since; R, h and C are constants
1eV = 1.6 x 10-19J
1MeV = 106eV
1MeV = 1.6 x 10-13J
√It was discovered that the energy increases from the nucleus of an atom towards the highest energy level.
√The energy increases from first shell toward seventh shell.
√From seventh shell the energy is zero or the energy at infinite is zero. Then the energy from infinite which is zero towards the lowest shell
decreases and results to be negative value of energy.
n∞ = 0.00ev
n7 = -0.15ev
n6 = -0.21ev
n5 = -0.54ev
n4 = -0.84ev
n3 = -1.50ev
n2 = -3.40ev
n1 = -13.60ev
E = E2 – E1
= E – En
= 0 – En
= 0 –13.6eV
n²
E = - 13.6eV
n²
The negative value occurs because the energy of infinite is zero. Since energy increase from nuclear towards highest energy level result the energy of each shell.
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The wave length of H-spectrum determined is applied to find frequency and energy of the H- spectrum.
The wave number is inversely proportional to the square of the energy level differences ( ∆n²).
ṽ∝1
∆n²
1 = RH ∆n²
ṽ
ṽ = [ 1/n₁² - 1/n₁²]RH
Since; ṽ = 1/λ
1/λ = RH[ 1/n₁² - 1/n₂² ]………………. (3)
Where;
λ= wave length
n₁ = lowest energy level
n₂ = highest energy level
RH = Reydberg constant
RH is 1.09 x 10^7 m-¹
The value of n₁ and n₂ for H-spectrum can be obtained if given number of line and number of series. The value n₁ is equal number of series given. But the value n₂ is equal to the number of line plus the number of series.
Problem # 03;
Obtain the values of the wavelength and the corresponding energy, for n₁ and n₂ of the third line in Balmer’s series.
Solution;
Number of line = Third line
Number of series = Balmer’s series
n₁ = number of series = 2
n₂ = number of line + Number of series = 3 + 2
n = 5.
But;
1/λ = RH[ 1/n₁² - 1/n₂²]
=1.09 X 10^7m-¹[1/2²-1/5²]
=1.09 X 10^7m-¹[1/4-1/25]
= 1.09 X 10^7m-¹[21/100]
λ = 4.45 x 10^-7m of third line
Recall from eqn
E = hc/λ
= 6.3E-3 Js x 3.0E-8ms-¹
4.45E-7m
E=4.25 x 10^-19J
.: the energy of balmer series above = 4.25×10^-19J
Also;
f = c
λ
f = 3.0 × 10^8ms-¹
4.45 x 10^-7m
f = 6.74 x 10¹⁴s-¹
f = 6.74 x 10¹⁴Hz
.: the frequency should be 6.74 × 10¹⁴Hz
Transition Energy
Is the energy required to shift an electron from one shell to another.
The energy required to
shift electrons from one shell to another should be equal to the energy difference between the respective shells.
The energy difference between the lowest energy level E₁ and the highest energy level E₂ is obtained by the expression;-
E = E₂– E₁……………… (1)
The trend of transition of electrons is shown as;
E = E₂ – E₁
Transition takes place from E₁ to E₂ exactly.
E > E₂ – E₁
Transition take place from E₁ toward beyond E₂
E < E₂ – E₁ – Transition take place from E₁ and hang between E₁ and E₂
But if electrons gain enough energy which is equal to the ionization energy results the
electron to jump completely from ground state to infinite. These electrons cannot return back instead leave an atom ionized positively. The energy supplied to the atom results it to be excited and electron leave completely from the ground state to infinite. The energy supplied to the electrons is used as ionization energy and as kinetic energy.
The ionization energy of electrons is equal to the amount of energy associated by electron in the shell or quantum number where it belongs.
The expression below is used to find the energy associated by electrons or energy of that shell.
1 = RH
λ
Let; n₁ = n
n₂ = ∞
1 = RH[ 1 - 1 ]
λ n² ∞²
1 = RH[ 1 ] since; 1 = 0
λ n² ∞²
1 = RH
λ n²............................(i)
E = hc
λ
E = 1
hc λ............................(ii)
Substitute eqn (ii) into eqn (i).
E = RHhc
n²
E = 13.6eV
n²
1eV = 1.6 x 10-19J
1MeV = 106eV
1MeV = 1.6 x 10-13J
√It was discovered that the energy increases from the nucleus of an atom towards the highest energy level.
√The energy increases from first shell toward seventh shell.
√From seventh shell the energy is zero or the energy at infinite is zero. Then the energy from infinite which is zero towards the lowest shell
decreases and results to be negative value of energy.
n∞ = 0.00ev
n7 = -0.15ev
n6 = -0.21ev
n5 = -0.54ev
n4 = -0.84ev
n3 = -1.50ev
n2 = -3.40ev
n1 = -13.60ev
E = E2 – E1
= E – En
= 0 – En
= 0 –13.6eV
n²
E = - 13.6eV
n²
The negative value occurs because the energy of infinite is zero. Since energy increase from nuclear towards highest energy level result the energy of each shell.
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