Class 12 Exam > Class 12 Questions > Light from a hydrogen discharge tube is incid...
Start Learning for Free
Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:
- a)-4.2V
- b)-9.4 V
- c)-17.8 V
- d)+9.4 V
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Light from a hydrogen discharge tube is incident on the cathode of a p...
Explanation:
When light is incident on the cathode of a photoelectric cell, electrons are emitted from the surface of the cathode due to the photoelectric effect. These emitted electrons constitute the photocurrent.
The work function of the cathode surface is the minimum energy required to remove an electron from the surface of the cathode. In this case, the work function is given as 4.2 eV.
The voltage of the anode relative to the cathode determines the direction and intensity of the electric field in the photoelectric cell. It affects the motion of the emitted electrons and can control the photocurrent.
In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be such that it prevents the emitted electrons from reaching the anode. This can be achieved by creating an electric field that opposes the motion of the electrons.
To calculate the voltage required, we can use the equation:
V = E/q
Where V is the voltage, E is the energy, and q is the charge of an electron.
The energy of an electron can be calculated using the equation:
E = hf - φ
Where E is the energy, h is Planck's constant, f is the frequency of the incident light, and φ is the work function.
Since we want to reduce the photocurrent to zero, the energy of the emitted electrons should be equal to or less than zero. This means that the energy of the incident light should be equal to or greater than the work function:
hf ≥ φ
Now we can substitute the energy of the incident light into the voltage equation:
V = (hf - φ)/q
Since hf ≥ φ, the energy term in the numerator is positive. Therefore, to create a negative voltage that opposes the motion of the electrons, we need to make the voltage negative relative to the cathode. This means that the anode must have a lower voltage than the cathode.
The correct answer is option B, -9.4 V, because it is a negative voltage relative to the cathode.
When light is incident on the cathode of a photoelectric cell, electrons are emitted from the surface of the cathode due to the photoelectric effect. These emitted electrons constitute the photocurrent.
The work function of the cathode surface is the minimum energy required to remove an electron from the surface of the cathode. In this case, the work function is given as 4.2 eV.
The voltage of the anode relative to the cathode determines the direction and intensity of the electric field in the photoelectric cell. It affects the motion of the emitted electrons and can control the photocurrent.
In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be such that it prevents the emitted electrons from reaching the anode. This can be achieved by creating an electric field that opposes the motion of the electrons.
To calculate the voltage required, we can use the equation:
V = E/q
Where V is the voltage, E is the energy, and q is the charge of an electron.
The energy of an electron can be calculated using the equation:
E = hf - φ
Where E is the energy, h is Planck's constant, f is the frequency of the incident light, and φ is the work function.
Since we want to reduce the photocurrent to zero, the energy of the emitted electrons should be equal to or less than zero. This means that the energy of the incident light should be equal to or greater than the work function:
hf ≥ φ
Now we can substitute the energy of the incident light into the voltage equation:
V = (hf - φ)/q
Since hf ≥ φ, the energy term in the numerator is positive. Therefore, to create a negative voltage that opposes the motion of the electrons, we need to make the voltage negative relative to the cathode. This means that the anode must have a lower voltage than the cathode.
The correct answer is option B, -9.4 V, because it is a negative voltage relative to the cathode.
|
Explore Courses for Class 12 exam
|
|
Similar Class 12 Doubts
Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect answer is option 'B'. Can you explain this answer?
Question Description
Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect 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 Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect 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 Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect answer is option 'B'. Can you explain this answer?.
Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect 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 Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect 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 Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect 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 Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photocurrent to zero, the voltage of the anode relative to the cathode must be made:a)-4.2Vb)-9.4 Vc)-17.8 Vd)+9.4 VCorrect 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