The wavelength of light emitted in a transition from the first excited state in which the ionization energy is 126 kJ/mol to the ground state in which the ionization energy is 412 kJ/mol should be calculated using the concept of Bohr’s theory. Concept Introduction: The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light. Based on electrostatic interaction and law of motion, Bohr derived the following equation. E n = − 2 .18 × 10 − 18 J ( 1 n 2 ) where n gets an integer values such as n = 1, 2, 3 and so on. This is the energy of electron in n th orbital. The electrons are excited thermally when the light is used by an object. As a result, an emission spectrum comes. Line spectra consist of light only at specific, discrete wavelengths. In emission, the electron returns to a lower energy state from n f (the i and f subscripts denote the initial and final energy states). In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state. The difference in the energies between the initial and final states is ΔE = E f − E i This transition results in the photon’s emission with frequency v and energy hv . The following equation is resulted. ΔE = hν = − 2 .18 × 10 − 18 J ( 1 n f 2 − 1 n i 2 ) When n i > n f , a photon is emitted. The term in parentheses is positive, making ΔE negative . As a result, energy is lost to the surroundings. When n i < n f , a photon is absorbed. The term in parentheses is negative, so ΔE is positive. As a result, energy is absorbed from the surroundings. To find: The wavelength of light emitted in a transition from the first excited state in which the ionization energy is 126 kJ/mol to the ground state in which the ionization energy is 412 kJ/mol
The wavelength of light emitted in a transition from the first excited state in which the ionization energy is 126 kJ/mol to the ground state in which the ionization energy is 412 kJ/mol should be calculated using the concept of Bohr’s theory. Concept Introduction: The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light. Based on electrostatic interaction and law of motion, Bohr derived the following equation. E n = − 2 .18 × 10 − 18 J ( 1 n 2 ) where n gets an integer values such as n = 1, 2, 3 and so on. This is the energy of electron in n th orbital. The electrons are excited thermally when the light is used by an object. As a result, an emission spectrum comes. Line spectra consist of light only at specific, discrete wavelengths. In emission, the electron returns to a lower energy state from n f (the i and f subscripts denote the initial and final energy states). In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state. The difference in the energies between the initial and final states is ΔE = E f − E i This transition results in the photon’s emission with frequency v and energy hv . The following equation is resulted. ΔE = hν = − 2 .18 × 10 − 18 J ( 1 n f 2 − 1 n i 2 ) When n i > n f , a photon is emitted. The term in parentheses is positive, making ΔE negative . As a result, energy is lost to the surroundings. When n i < n f , a photon is absorbed. The term in parentheses is negative, so ΔE is positive. As a result, energy is absorbed from the surroundings. To find: The wavelength of light emitted in a transition from the first excited state in which the ionization energy is 126 kJ/mol to the ground state in which the ionization energy is 412 kJ/mol
Solution Summary: The author explains that the wavelength of light emitted in a transition from the first excited state to the ground state should be calculated using the concept of Bohr's theory.
The wavelength of light emitted in a transition from the first excited state in which the ionization energy is 126 kJ/mol to the ground state in which the ionization energy is 412 kJ/mol should be calculated using the concept of Bohr’s theory.
Concept Introduction:
The emission of radiation given by an energized hydrogen atom to the electron falling from a higher-energy orbit to a lower orbit give a quantum of energy in the form of light. Based on electrostatic interaction and law of motion, Bohr derived the following equation.
En=−2.18 × 10−18 J (1n2)
where n gets an integer values such as n = 1, 2, 3 and so on. This is the energy of electron in nth orbital.
The electrons are excited thermally when the light is used by an object. As a result, an emission spectrum comes. Line spectra consist of light only at specific, discrete wavelengths. In emission, the electron returns to a lower energy state from nf (the i and f subscripts denote the initial and final energy states). In most cases, the lower energy state corresponds to the ground state but it may be any energy state which is lower than the initial excited state. The difference in the energies between the initial and final states is
ΔE = Ef− Ei
This transition results in the photon’s emission with frequency v and energy hv. The following equation is resulted.
ΔE = hν =−2.18 × 10−18 J (1nf2−1ni2)
When ni > nf, a photon is emitted. The term in parentheses is positive, making ΔE negative. As a result, energy is lost to the surroundings. When ni < nf, a photon is absorbed. The term in parentheses is negative, so ΔE is positive. As a result, energy is absorbed from the surroundings.
To find: The wavelength of light emitted in a transition from the first excited state in which the ionization energy is 126 kJ/mol to the ground state in which the ionization energy is 412 kJ/mol
Identify the starting material in the following reaction. Click the "draw structure" button to launch the
drawing utility.
draw structure ...
[1] 0 3
C10H18
[2] CH3SCH3
H
In an equilibrium mixture of the formation of ammonia from nitrogen and hydrogen, it is found that
PNH3 = 0.147 atm, PN2 = 1.41 atm and Pн2 = 6.00 atm. Evaluate Kp and Kc at 500 °C.
2 NH3 (g) N2 (g) + 3 H₂ (g)
K₂ = (PN2)(PH2)³ = (1.41) (6.00)³ = 1.41 x 104
What alkene or alkyne yields the following products after oxidative cleavage with ozone? Click the
"draw structure" button to launch the drawing utility.
and two equivalents of CH2=O
draw structure ...
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