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The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]
- a)1.54 x 1015 s -1
- b)1.03 x 1015 s -1
- c)3.08 x 1015 s -1
- d)2.00 x 1015 s -1
Correct answer is option 'C'. Can you explain this answer?
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The frequency of radiation emitted when the electron falls from n = 4 ...
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The frequency of radiation emitted when the electron falls from n = 4 ...
The frequency of radiation emitted when an electron falls from a higher energy level to a lower energy level in a hydrogen atom can be calculated using the formula:
ΔE = E_final - E_initial = -13.6 eV * (1/n_final^2 - 1/n_initial^2)
Where:
ΔE is the change in energy between the final and initial states
E_final is the energy of the final state
E_initial is the energy of the initial state
n_final is the principal quantum number of the final state
n_initial is the principal quantum number of the initial state
In this case, the initial principal quantum number is n_initial = 4 and the final principal quantum number is n_final = 1.
Plugging these values into the formula:
ΔE = -13.6 eV * (1/1^2 - 1/4^2)
= -13.6 eV * (1 - 1/16)
= -13.6 eV * (15/16)
= -13.6 eV * 0.9375
= -12.75 eV
Since the electron is falling to a lower energy level, the change in energy is negative.
To convert the change in energy to frequency, we can use the formula:
ΔE = h * f
Where:
h is Planck's constant (6.626 x 10^-34 J·s)
f is the frequency
Rearranging the formula to solve for f:
f = ΔE / h
Plugging in the values:
f = (-12.75 eV * 1.602 x 10^-19 J/eV) / (6.626 x 10^-34 J·s)
= -20.43 x 10^14 Hz
The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom is approximately 20.43 x 10^14 Hz.
ΔE = E_final - E_initial = -13.6 eV * (1/n_final^2 - 1/n_initial^2)
Where:
ΔE is the change in energy between the final and initial states
E_final is the energy of the final state
E_initial is the energy of the initial state
n_final is the principal quantum number of the final state
n_initial is the principal quantum number of the initial state
In this case, the initial principal quantum number is n_initial = 4 and the final principal quantum number is n_final = 1.
Plugging these values into the formula:
ΔE = -13.6 eV * (1/1^2 - 1/4^2)
= -13.6 eV * (1 - 1/16)
= -13.6 eV * (15/16)
= -13.6 eV * 0.9375
= -12.75 eV
Since the electron is falling to a lower energy level, the change in energy is negative.
To convert the change in energy to frequency, we can use the formula:
ΔE = h * f
Where:
h is Planck's constant (6.626 x 10^-34 J·s)
f is the frequency
Rearranging the formula to solve for f:
f = ΔE / h
Plugging in the values:
f = (-12.75 eV * 1.602 x 10^-19 J/eV) / (6.626 x 10^-34 J·s)
= -20.43 x 10^14 Hz
The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom is approximately 20.43 x 10^14 Hz.
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The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]a)1.54 x 1015 s -1b)1.03 x 1015 s -1c)3.08 x 1015 s -1d)2.00 x 1015 s -1Correct answer is option 'C'. Can you explain this answer?
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The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]a)1.54 x 1015 s -1b)1.03 x 1015 s -1c)3.08 x 1015 s -1d)2.00 x 1015 s -1Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]a)1.54 x 1015 s -1b)1.03 x 1015 s -1c)3.08 x 1015 s -1d)2.00 x 1015 s -1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]a)1.54 x 1015 s -1b)1.03 x 1015 s -1c)3.08 x 1015 s -1d)2.00 x 1015 s -1Correct answer is option 'C'. Can you explain this answer?.
The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]a)1.54 x 1015 s -1b)1.03 x 1015 s -1c)3.08 x 1015 s -1d)2.00 x 1015 s -1Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]a)1.54 x 1015 s -1b)1.03 x 1015 s -1c)3.08 x 1015 s -1d)2.00 x 1015 s -1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H=2.18 ×10–18J atom–1and h = 6.625 × 10–34 J s ) [2004]a)1.54 x 1015 s -1b)1.03 x 1015 s -1c)3.08 x 1015 s -1d)2.00 x 1015 s -1Correct answer is option 'C'. Can you explain this answer?.
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