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The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42 x 10–12 erg and –2.41 x 10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.
Correct answer is '6604'. Can you explain this answer?
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The energy of electron in the second and third Bohr orbit of the hydro...
The energy of an electron in the nth Bohr orbit of a hydrogen atom is given by the formula:
E = -13.6 eV / n^2
Where E is the energy, n is the principal quantum number (the Bohr orbit number), and -13.6 eV is the ionization energy of hydrogen.
For the second Bohr orbit (n = 2):
E = -13.6 eV / 2^2
E = -13.6 eV / 4
E = -3.4 eV
For the third Bohr orbit (n = 3):
E = -13.6 eV / 3^2
E = -13.6 eV / 9
E = -1.51 eV
Therefore, the energy of an electron in the second Bohr orbit is -3.4 eV, and the energy of an electron in the third Bohr orbit is -1.51 eV.
E = -13.6 eV / n^2
Where E is the energy, n is the principal quantum number (the Bohr orbit number), and -13.6 eV is the ionization energy of hydrogen.
For the second Bohr orbit (n = 2):
E = -13.6 eV / 2^2
E = -13.6 eV / 4
E = -3.4 eV
For the third Bohr orbit (n = 3):
E = -13.6 eV / 3^2
E = -13.6 eV / 9
E = -1.51 eV
Therefore, the energy of an electron in the second Bohr orbit is -3.4 eV, and the energy of an electron in the third Bohr orbit is -1.51 eV.
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The energy of electron in the second and third Bohr orbit of the hydro...
E3 – E2 = hv = hc/λ – 2.41 ´ 10–12 – (– 5.42 ´ 10–12)
= 6.626 x 10-27 x 3 x 1010/ λ
∴ λ = 6.626 x 10-27 x 3 x 1010 / 3.01 x 10-12
= 6.604 x 10–5 cm = 6.604 x 10–5 x 108
= 6604 Å
= 6.626 x 10-27 x 3 x 1010/ λ
∴ λ = 6.626 x 10-27 x 3 x 1010 / 3.01 x 10-12
= 6.604 x 10–5 cm = 6.604 x 10–5 x 108
= 6604 Å
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The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42x10–12erg and –2.41x10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.Correct answer is '6604'. Can you explain this answer?
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The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42x10–12erg and –2.41x10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.Correct answer is '6604'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42x10–12erg and –2.41x10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.Correct answer is '6604'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42x10–12erg and –2.41x10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.Correct answer is '6604'. Can you explain this answer?.
The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42x10–12erg and –2.41x10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.Correct answer is '6604'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42x10–12erg and –2.41x10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.Correct answer is '6604'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.42x10–12erg and –2.41x10–12erg, respectively. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.Correct answer is '6604'. Can you explain this answer?.
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