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A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Then
- a)I = 10 mA
- b)I=100 mA
- c)S1 = 0.775V
- d)S2 = 3.875V
Correct answer is option 'C,D'. Can you explain this answer?
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Verified Answer
A beam of light of wavelength 400 nm and power 1.55 mQ is directed at ...
Wavelength of incident light λ= 400nm
P = 1.55 mW
Energy of one photon
Number of photons falling per second
Number of electrons ejected is 10% of number of photon falling per second .
N = 0.1 n
Current constipated by flow of ejected electrons
Kinetic theory of ejected electron = Photon - Work function Energy of incident
Where λ = 400nm is reduced to λ = 200nm , the KE increases by 5 times.
Subtracting equation (A) from (B)
Stopping potential
Stopping potential
P = 1.55 mW
Energy of one photon
Number of photons falling per second
Number of electrons ejected is 10% of number of photon falling per second .
N = 0.1 n
Current constipated by flow of ejected electrons
Kinetic theory of ejected electron = Photon - Work function Energy of incident
Where λ = 400nm is reduced to λ = 200nm , the KE increases by 5 times.
Subtracting equation (A) from (B)
Stopping potential
Stopping potential
Most Upvoted Answer
A beam of light of wavelength 400 nm and power 1.55 mQ is directed at ...
Given:
- Wavelength of light incident on the photoelectric cell, λ1 = 400 nm
- Power of the incident light, P1 = 1.55 mW = 1.55 × 10^-3 W
- Energy of a photon, hν1 = hc/λ1, where h = 1240 eV·nm and c = 3 × 10^8 m/s
- Percentage of current due to electrons, I = 10%
- New wavelength of light, λ2 = 200 nm
To find:
- Current due to electrons, I
- Stopping potentials for the two wavelengths, S1 and S2
Solution:
1. Calculation of Energy of a Photon:
- Energy of a photon is given by the equation E = hν, where E is the energy, h is the Planck's constant, and ν is the frequency of the photon.
- Since frequency is related to wavelength by the equation c = λν, where c is the speed of light, we can rewrite the energy equation as E = hc/λ.
- Substituting the given values, we can calculate the energy of a photon for both λ1 and λ2.
2. Calculation of Current:
- The current due to electrons is given as a percentage of the total current.
- We are given that only 10% of the current is due to electrons, so we can calculate the current using the given power and the energy of the photons.
- The power of the incident light is the rate at which energy is delivered to the photoelectric cell.
- Since the energy of a photon is given by E = P/ν, where P is the power and ν is the frequency, we can calculate the number of photons incident on the cathode per second.
- The number of electrons emitted per second is equal to the number of incident photons multiplied by the percentage of current due to electrons.
- Finally, the current is given by the equation I = q/t, where q is the charge and t is the time.
3. Calculation of Stopping Potentials:
- The stopping potential for a photoelectric cell is the minimum potential difference required to stop the flow of electrons.
- The kinetic energy of the emitted electrons is given by the equation K.E. = eV, where e is the charge of an electron and V is the stopping potential.
- We are given that the kinetic energy of the electrons increases by a factor of 5 when the wavelength is reduced to 200 nm.
- We can use this information to calculate the stopping potentials for the two wavelengths using the equation K.E. = 1/2mv^2.
Summary:
- The current due to electrons is calculated by considering the power of the incident light and the energy of the photons.
- The stopping potentials for the two wavelengths can be calculated using the change in kinetic energy of the emitted electrons.
- Wavelength of light incident on the photoelectric cell, λ1 = 400 nm
- Power of the incident light, P1 = 1.55 mW = 1.55 × 10^-3 W
- Energy of a photon, hν1 = hc/λ1, where h = 1240 eV·nm and c = 3 × 10^8 m/s
- Percentage of current due to electrons, I = 10%
- New wavelength of light, λ2 = 200 nm
To find:
- Current due to electrons, I
- Stopping potentials for the two wavelengths, S1 and S2
Solution:
1. Calculation of Energy of a Photon:
- Energy of a photon is given by the equation E = hν, where E is the energy, h is the Planck's constant, and ν is the frequency of the photon.
- Since frequency is related to wavelength by the equation c = λν, where c is the speed of light, we can rewrite the energy equation as E = hc/λ.
- Substituting the given values, we can calculate the energy of a photon for both λ1 and λ2.
2. Calculation of Current:
- The current due to electrons is given as a percentage of the total current.
- We are given that only 10% of the current is due to electrons, so we can calculate the current using the given power and the energy of the photons.
- The power of the incident light is the rate at which energy is delivered to the photoelectric cell.
- Since the energy of a photon is given by E = P/ν, where P is the power and ν is the frequency, we can calculate the number of photons incident on the cathode per second.
- The number of electrons emitted per second is equal to the number of incident photons multiplied by the percentage of current due to electrons.
- Finally, the current is given by the equation I = q/t, where q is the charge and t is the time.
3. Calculation of Stopping Potentials:
- The stopping potential for a photoelectric cell is the minimum potential difference required to stop the flow of electrons.
- The kinetic energy of the emitted electrons is given by the equation K.E. = eV, where e is the charge of an electron and V is the stopping potential.
- We are given that the kinetic energy of the electrons increases by a factor of 5 when the wavelength is reduced to 200 nm.
- We can use this information to calculate the stopping potentials for the two wavelengths using the equation K.E. = 1/2mv^2.
Summary:
- The current due to electrons is calculated by considering the power of the incident light and the energy of the photons.
- The stopping potentials for the two wavelengths can be calculated using the change in kinetic energy of the emitted electrons.
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A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer?
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A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer?.
A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer?.
Solutions for A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer?, a detailed solution for A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? has been provided alongside types of A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A beam of light of wavelength 400 nm and power 1.55 mQ is directed at the cathode of a photoelectric cell, (given he = 1240 ev). If only 10% of the current due to these electrons is i . If the wavelength o f light is now reduced to 200 nm, keeping its power the same, the kinetic energy of the electrons is found to increase by a factor of 5. The stopping potentials for the two wavelength are S1 and S2. Thena)I = 10 mAb)I=100 mAc)S1= 0.775Vd)S2 = 3.875VCorrect answer is option 'C,D'. Can you explain this answer? tests, examples and also practice GATE tests.
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