NEET Exam > NEET Questions > Radiation of wavelength 2500 am strong is inc...
Start Learning for Free
Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is?
Most Upvoted Answer
Radiation of wavelength 2500 am strong is incident on the metal plate ...
Key Points:
- Incident radiation has a wavelength of 2500 am.
- The work function of the metal plate is 3.5 eV.
- We need to calculate the potential required to stop the fastest emitted electrons.
Explanation:
1. Understanding the Work Function:
The work function of a metal is the minimum amount of energy required to remove an electron from its surface. When radiation of sufficient energy is incident on a metal surface, electrons can be emitted through the photoelectric effect.
2. Relation between Energy and Wavelength:
The energy of a photon is related to its wavelength by the equation: E = hc/λ, where E is the energy, h is the Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
3. Calculating the Energy of Incident Radiation:
Since the wavelength of the incident radiation is given as 2500 am (attometer), we need to convert it to meters. 1 am = 10^-18 m.
Wavelength in meters (λ) = 2500 am x 10^-18 m/am = 2.5 x 10^-15 m
Using the energy-wavelength relation, we can calculate the energy (E) of the incident radiation:
E = hc/λ = (6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (2.5 x 10^-15 m) = 7.9512 x 10^-19 J
4. Determining the Maximum Kinetic Energy of Electrons:
The maximum kinetic energy (KEmax) of emitted electrons is given by the difference between the energy of the incident radiation and the work function of the metal:
KEmax = E - work function
The work function is given as 3.5 eV. To convert eV to Joules, we use the conversion factor: 1 eV = 1.6 x 10^-19 J.
Work function in Joules = 3.5 eV x (1.6 x 10^-19 J/eV) = 5.6 x 10^-19 J
KEmax = 7.9512 x 10^-19 J - 5.6 x 10^-19 J = 2.3512 x 10^-19 J
5. Finding the Required Potential:
The potential required to stop the fastest emitted electrons is equal to their maximum kinetic energy. Using the equation:
KEmax = eV
where e is the elementary charge (1.6 x 10^-19 C) and V is the potential, we can solve for V:
V = KEmax / e = (2.3512 x 10^-19 J) / (1.6 x 10^-19 C) = 1.4695 V
Thus, the potential required to stop the fastest emitted electrons is approximately 1.47 V.
- Incident radiation has a wavelength of 2500 am.
- The work function of the metal plate is 3.5 eV.
- We need to calculate the potential required to stop the fastest emitted electrons.
Explanation:
1. Understanding the Work Function:
The work function of a metal is the minimum amount of energy required to remove an electron from its surface. When radiation of sufficient energy is incident on a metal surface, electrons can be emitted through the photoelectric effect.
2. Relation between Energy and Wavelength:
The energy of a photon is related to its wavelength by the equation: E = hc/λ, where E is the energy, h is the Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
3. Calculating the Energy of Incident Radiation:
Since the wavelength of the incident radiation is given as 2500 am (attometer), we need to convert it to meters. 1 am = 10^-18 m.
Wavelength in meters (λ) = 2500 am x 10^-18 m/am = 2.5 x 10^-15 m
Using the energy-wavelength relation, we can calculate the energy (E) of the incident radiation:
E = hc/λ = (6.626 x 10^-34 J·s) x (3 x 10^8 m/s) / (2.5 x 10^-15 m) = 7.9512 x 10^-19 J
4. Determining the Maximum Kinetic Energy of Electrons:
The maximum kinetic energy (KEmax) of emitted electrons is given by the difference between the energy of the incident radiation and the work function of the metal:
KEmax = E - work function
The work function is given as 3.5 eV. To convert eV to Joules, we use the conversion factor: 1 eV = 1.6 x 10^-19 J.
Work function in Joules = 3.5 eV x (1.6 x 10^-19 J/eV) = 5.6 x 10^-19 J
KEmax = 7.9512 x 10^-19 J - 5.6 x 10^-19 J = 2.3512 x 10^-19 J
5. Finding the Required Potential:
The potential required to stop the fastest emitted electrons is equal to their maximum kinetic energy. Using the equation:
KEmax = eV
where e is the elementary charge (1.6 x 10^-19 C) and V is the potential, we can solve for V:
V = KEmax / e = (2.3512 x 10^-19 J) / (1.6 x 10^-19 C) = 1.4695 V
Thus, the potential required to stop the fastest emitted electrons is approximately 1.47 V.
Community Answer
Radiation of wavelength 2500 am strong is incident on the metal plate ...
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is?
Question Description
Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is?.
Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is?.
Solutions for Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? defined & explained in the simplest way possible. Besides giving the explanation of
Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is?, a detailed solution for Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? has been provided alongside types of Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? theory, EduRev gives you an
ample number of questions to practice Radiation of wavelength 2500 am strong is incident on the metal plate whose work function is 3.5 eV then the potential required to stop the fastest Photon electrons emitted by the surface is? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup