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The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.?
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The frequency of radiation emitted in the electron fall from n equals ...
Understanding Electron Transitions in Hydrogen
In a hydrogen atom, when an electron transitions from a higher energy level (n=4) to a lower one (n=1), it emits radiation. The energy difference between these levels can be calculated using the formula:
Energy of emitted photon:
The energy difference (ΔE) between two levels is given by:
ΔE = E_n1 - E_n4
Where E_n is the energy of the electron in the nth level, calculated as:
E_n = - (Ionization Energy) / n²
Calculating Energy Levels:
- For n=1:
E_1 = - (2.18 x 10^-18 J) / (1²) = -2.18 x 10^-18 J
- For n=4:
E_4 = - (2.18 x 10^-18 J) / (4²) = -2.18 x 10^-18 J / 16 = -0.13625 x 10^-18 J
Determining ΔE:
ΔE = E_1 - E_4
ΔE = (-2.18 x 10^-18) - (-0.13625 x 10^-18)
ΔE = -2.18 x 10^-18 + 0.13625 x 10^-18 = -2.04375 x 10^-18 J
Calculating the Frequency of Emitted Radiation:
Using the relationship between energy and frequency:
E = h * f
Where h is Planck’s constant (6.626 x 10^-34 J·s).
Rearranging gives:
f = ΔE / h
Substituting the values:
f = (-2.04375 x 10^-18 J) / (6.626 x 10^-34 J·s)
This will yield the frequency of the emitted radiation when the electron falls from n=4 to n=1.
Conclusion:
By following these steps, one can determine the frequency of radiation associated with electron transitions in hydrogen, which is fundamental in quantum mechanics and spectroscopy.
In a hydrogen atom, when an electron transitions from a higher energy level (n=4) to a lower one (n=1), it emits radiation. The energy difference between these levels can be calculated using the formula:
Energy of emitted photon:
The energy difference (ΔE) between two levels is given by:
ΔE = E_n1 - E_n4
Where E_n is the energy of the electron in the nth level, calculated as:
E_n = - (Ionization Energy) / n²
Calculating Energy Levels:
- For n=1:
E_1 = - (2.18 x 10^-18 J) / (1²) = -2.18 x 10^-18 J
- For n=4:
E_4 = - (2.18 x 10^-18 J) / (4²) = -2.18 x 10^-18 J / 16 = -0.13625 x 10^-18 J
Determining ΔE:
ΔE = E_1 - E_4
ΔE = (-2.18 x 10^-18) - (-0.13625 x 10^-18)
ΔE = -2.18 x 10^-18 + 0.13625 x 10^-18 = -2.04375 x 10^-18 J
Calculating the Frequency of Emitted Radiation:
Using the relationship between energy and frequency:
E = h * f
Where h is Planck’s constant (6.626 x 10^-34 J·s).
Rearranging gives:
f = ΔE / h
Substituting the values:
f = (-2.04375 x 10^-18 J) / (6.626 x 10^-34 J·s)
This will yield the frequency of the emitted radiation when the electron falls from n=4 to n=1.
Conclusion:
By following these steps, one can determine the frequency of radiation associated with electron transitions in hydrogen, which is fundamental in quantum mechanics and spectroscopy.
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The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.?
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The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.?.
The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.?.
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The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.?, a detailed solution for The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.? has been provided alongside types of The frequency of radiation emitted in the electron fall from n equals to 4 to n equals to 1 in hydrogen atom will be given ionization energy of hydrogen equals to 2.18 into 10 to the power minus 18 joule per atom. And h equals to 6.626 into 10 to the power minus 34 joule second.? theory, EduRev gives you an
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