Class 11 Exam > Class 11 Questions > The frequency of radiation emitted when the e...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The frequency of radiation emitted when the electron falls from n = 4 ...
Most Upvoted Answer
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.
Free Test
FREE
| Start Free Test |
Community Answer
The frequency of radiation emitted when the electron falls from n = 4 ...
Plz solve
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup