Class 12 Exam > Class 12 Questions > The photoelectric work function for a metalsu...
Start Learning for Free
The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]
- a)4125 Å
- b)3000 Å
- c)6000 Å
- d)2062.5 Å
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The photoelectric work function for a metalsurface is 4.125 eV. The cu...
Let λ0 be cut off wavelength.
Most Upvoted Answer
The photoelectric work function for a metalsurface is 4.125 eV. The cu...
Explanation:
Photoelectric Work Function:
- The photoelectric work function is the minimum amount of energy required to remove an electron from the surface of a material.
- In this case, the photoelectric work function is given as 4.125 eV.
Relationship between Work Function and Cut-off Wavelength:
- The cut-off wavelength, λ, can be calculated using the formula:
λ = hc / Φ
where h is the Planck's constant (4.135 x 10^-15 eV s), c is the speed of light (3 x 10^8 m/s), and Φ is the work function.
Calculation:
- Substituting the given values into the formula:
λ = (4.135 x 10^-15 eV s * 3 x 10^8 m/s) / 4.125 eV
λ ≈ (1.2405 x 10^-6 m) / 4.125
λ ≈ 3 x 10^-7 m
Conversion to Angstrom:
- 1 meter is equal to 10^10 Angstroms. Therefore, to convert the result to Angstrom:
λ = 3 x 10^-7 m * 10^10 Å/m
λ = 3000 Å
Therefore, the cut-off wavelength for the given metal surface is 3000 Å, which corresponds to option b.
Photoelectric Work Function:
- The photoelectric work function is the minimum amount of energy required to remove an electron from the surface of a material.
- In this case, the photoelectric work function is given as 4.125 eV.
Relationship between Work Function and Cut-off Wavelength:
- The cut-off wavelength, λ, can be calculated using the formula:
λ = hc / Φ
where h is the Planck's constant (4.135 x 10^-15 eV s), c is the speed of light (3 x 10^8 m/s), and Φ is the work function.
Calculation:
- Substituting the given values into the formula:
λ = (4.135 x 10^-15 eV s * 3 x 10^8 m/s) / 4.125 eV
λ ≈ (1.2405 x 10^-6 m) / 4.125
λ ≈ 3 x 10^-7 m
Conversion to Angstrom:
- 1 meter is equal to 10^10 Angstroms. Therefore, to convert the result to Angstrom:
λ = 3 x 10^-7 m * 10^10 Å/m
λ = 3000 Å
Therefore, the cut-off wavelength for the given metal surface is 3000 Å, which corresponds to option b.
|
Explore Courses for Class 12 exam
|
|
Similar Class 12 Doubts
The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer?
Question Description
The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer?.
The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer?.
Solutions for The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The photoelectric work function for a metalsurface is 4.125 eV. The cut off wavelength forthis surface is [1999]a)4125 Åb)3000 Åc)6000 Åd)2062.5 ÅCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
|
|
Explore Courses for Class 12 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