JEE Exam > JEE Questions > The work function of Potassium is 2.0 eV. Whe...
Start Learning for Free
The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons is
- a)0.75V
- b)1.75V
- c)2.5V
- d)3.75V
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
The work function of Potassium is 2.0 eV. When it is illuminated by li...
Free Test
FREE
| Start Free Test |
Community Answer
The work function of Potassium is 2.0 eV. When it is illuminated by li...
To determine the stopping potential of photoelectrons emitted when Potassium is illuminated by light of wavelength 3300 A0, we need to use the concept of the photoelectric effect and the equation for the stopping potential.
The photoelectric effect states that when light of a certain frequency or wavelength is incident on a metal surface, electrons are emitted from the surface. The energy of the incident photons must be greater than or equal to the work function of the metal for this emission to occur.
Given data:
- Work function of Potassium (φ) = 2.0 eV
- Wavelength of incident light (λ) = 3300 A0 (1 A0 = 10^-10 m)
1. Calculate the energy of the incident photons:
The energy (E) of a photon can be calculated using the equation:
E = hc/λ
where h is the Planck's constant (6.63 x 10^-34 J.s) and c is the speed of light (3 x 10^8 m/s).
E = (6.63 x 10^-34 J.s * 3 x 10^8 m/s) / (3300 x 10^-10 m)
E ≈ 6.02 x 10^-19 J
2. Convert the energy to electron volts (eV):
1 eV = 1.6 x 10^-19 J
E = (6.02 x 10^-19 J) / (1.6 x 10^-19 J/eV)
E ≈ 3.76 eV
3. Calculate the stopping potential (V):
The stopping potential (V) is the minimum potential difference required to stop the emitted photoelectrons from reaching the anode.
The stopping potential can be calculated using the equation:
eV = E - φ
where e is the elementary charge (1.6 x 10^-19 C).
eV = (3.76 eV) - (2.0 eV)
V ≈ 1.76 V
Therefore, the stopping potential of the photoelectrons emitted when Potassium is illuminated by light of wavelength 3300 A0 is approximately 1.76 V.
Option 'B' (1.75V) is the closest value to the calculated stopping potential.
The photoelectric effect states that when light of a certain frequency or wavelength is incident on a metal surface, electrons are emitted from the surface. The energy of the incident photons must be greater than or equal to the work function of the metal for this emission to occur.
Given data:
- Work function of Potassium (φ) = 2.0 eV
- Wavelength of incident light (λ) = 3300 A0 (1 A0 = 10^-10 m)
1. Calculate the energy of the incident photons:
The energy (E) of a photon can be calculated using the equation:
E = hc/λ
where h is the Planck's constant (6.63 x 10^-34 J.s) and c is the speed of light (3 x 10^8 m/s).
E = (6.63 x 10^-34 J.s * 3 x 10^8 m/s) / (3300 x 10^-10 m)
E ≈ 6.02 x 10^-19 J
2. Convert the energy to electron volts (eV):
1 eV = 1.6 x 10^-19 J
E = (6.02 x 10^-19 J) / (1.6 x 10^-19 J/eV)
E ≈ 3.76 eV
3. Calculate the stopping potential (V):
The stopping potential (V) is the minimum potential difference required to stop the emitted photoelectrons from reaching the anode.
The stopping potential can be calculated using the equation:
eV = E - φ
where e is the elementary charge (1.6 x 10^-19 C).
eV = (3.76 eV) - (2.0 eV)
V ≈ 1.76 V
Therefore, the stopping potential of the photoelectrons emitted when Potassium is illuminated by light of wavelength 3300 A0 is approximately 1.76 V.
Option 'B' (1.75V) is the closest value to the calculated stopping potential.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer?
Question Description
The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer?.
The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer?.
Solutions for The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE 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