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In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4 N due to the impact of the electrons. The value of n is __________. Mass of the electron me = 9 x 10-31 kg and 1.0 eV = 1.6 x 10-19 J.
Correct answer is '24.00'. Can you explain this answer?
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Given data:
- Power of incident light (P) = 200 W
- Work function of the cathode (ϕ) = 6.25 eV
- Threshold frequency (ν₀) = ?
- Efficiency of photoelectron emission (η) = 100%
- Potential difference between cathode and anode (V) = 500 V
- Force experienced by the anode due to electron impact (F) = n × 10⁻⁴ N
- Mass of the electron (mₑ) = 9 × 10⁻³¹ kg
- 1.0 eV = 1.6 × 10⁻¹⁹ J
Calculating the threshold frequency:
The work function (ϕ) represents the minimum energy required to remove an electron from the cathode. This energy is given by the equation:
ϕ = hν₀
Where h is the Planck's constant (6.63 × 10⁻³⁴ J·s) and ν₀ is the threshold frequency.
To convert the work function from electron volts (eV) to joules (J), we use the conversion factor:
1.0 eV = 1.6 × 10⁻¹⁹ J
Therefore, ϕ = 6.25 eV × (1.6 × 10⁻¹⁹ J/eV) = 1 × 10⁻¹⁸ J
Using the equation ϕ = hν₀, we can solve for ν₀:
ν₀ = ϕ / h = (1 × 10⁻¹⁸ J) / (6.63 × 10⁻³⁴ J·s) = 1.51 × 10¹⁵ Hz
Calculating the energy of incident photons:
The power (P) of the incident light is given by the equation:
P = nhν
Where n is the number of photons incident per second, h is the Planck's constant, and ν is the frequency of the incident light.
Rearranging the equation, we get:
n = P / (hν)
Substituting the given values, we have:
n = 200 W / (6.63 × 10⁻³⁴ J·s × 1.51 × 10¹⁵ Hz) ≈ 1.85 × 10²⁰ photons/s
Calculating the force experienced by the anode:
The force (F) experienced by the anode due to the impact of electrons can be calculated using the equation:
F = nqE
Where n is the number of electrons incident per second, q is the charge of an electron, and E is the electric field strength.
Since all emitted electrons are incident normally on the anode and absorbed, the number of electrons incident per second is equal to the number of photons incident per second, which is n = 1.85 × 10²⁰ electrons/s.
The charge of an electron (q) is -1.6 × 10⁻¹⁹ C.
Substituting the given values, we have:
n × 10⁻⁴ N = (1.85 × 10²⁰ electrons/s) × (-1.6 × 10⁻
- Power of incident light (P) = 200 W
- Work function of the cathode (ϕ) = 6.25 eV
- Threshold frequency (ν₀) = ?
- Efficiency of photoelectron emission (η) = 100%
- Potential difference between cathode and anode (V) = 500 V
- Force experienced by the anode due to electron impact (F) = n × 10⁻⁴ N
- Mass of the electron (mₑ) = 9 × 10⁻³¹ kg
- 1.0 eV = 1.6 × 10⁻¹⁹ J
Calculating the threshold frequency:
The work function (ϕ) represents the minimum energy required to remove an electron from the cathode. This energy is given by the equation:
ϕ = hν₀
Where h is the Planck's constant (6.63 × 10⁻³⁴ J·s) and ν₀ is the threshold frequency.
To convert the work function from electron volts (eV) to joules (J), we use the conversion factor:
1.0 eV = 1.6 × 10⁻¹⁹ J
Therefore, ϕ = 6.25 eV × (1.6 × 10⁻¹⁹ J/eV) = 1 × 10⁻¹⁸ J
Using the equation ϕ = hν₀, we can solve for ν₀:
ν₀ = ϕ / h = (1 × 10⁻¹⁸ J) / (6.63 × 10⁻³⁴ J·s) = 1.51 × 10¹⁵ Hz
Calculating the energy of incident photons:
The power (P) of the incident light is given by the equation:
P = nhν
Where n is the number of photons incident per second, h is the Planck's constant, and ν is the frequency of the incident light.
Rearranging the equation, we get:
n = P / (hν)
Substituting the given values, we have:
n = 200 W / (6.63 × 10⁻³⁴ J·s × 1.51 × 10¹⁵ Hz) ≈ 1.85 × 10²⁰ photons/s
Calculating the force experienced by the anode:
The force (F) experienced by the anode due to the impact of electrons can be calculated using the equation:
F = nqE
Where n is the number of electrons incident per second, q is the charge of an electron, and E is the electric field strength.
Since all emitted electrons are incident normally on the anode and absorbed, the number of electrons incident per second is equal to the number of photons incident per second, which is n = 1.85 × 10²⁰ electrons/s.
The charge of an electron (q) is -1.6 × 10⁻¹⁹ C.
Substituting the given values, we have:
n × 10⁻⁴ N = (1.85 × 10²⁰ electrons/s) × (-1.6 × 10⁻
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In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. Can you explain this answer?
Question Description
In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. 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 In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. Can you explain this answer?.
In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. 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 In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. Can you explain this answer?.
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ample number of questions to practice In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n x 10-4N due to the impact of the electrons. The value of n is __________. Mass of the electron me= 9 x10-31kg and 1.0 eV = 1.6 x10-19J.Correct answer is '24.00'. Can you explain this answer? tests, examples and also practice JEE tests.
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