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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?
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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 ...
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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?
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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?.
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