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Ultraviolet light of wavelength 350 nm and intensity 1.00 W/m2 is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s, 1 eV = 1.6 x 10-19 J)
Correct answer is between '8.7,8.9'. Can you explain this answer?
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Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is direct...
Energy of a photons is
Number of photons passing through unit surface per second is
The rate at which photoelectrons are emitted
Number of photons passing through unit surface per second is
The rate at which photoelectrons are emitted
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Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is direct...
Given data:
Wavelength of ultraviolet light, λ = 350 nm = 350 × 10⁻⁹ m
Intensity of light, I = 1.00 W/m²
Work function of metal surface, φ = 2.2 eV = 2.2 × 1.6 × 10⁻¹⁹ J
Percentage of photons producing photoelectrons, n = 0.50%
Area of metal surface, A = 1.00 cm² = 1.00 × 10⁻⁴ m²
Planck’s constant, h = 6.626 × 10⁻³⁴ J s
Speed of light, c = 3 × 10⁸ m/s
To find: Number of photoelectrons emitted per second
Solution:
1. Energy of photons:
The energy of a photon is given by the equation:
E = hc/λ
where h is Planck’s constant and c is the speed of light.
Substituting the given values, we get:
E = (6.626 × 10⁻³⁴ J s) × (3 × 10⁸ m/s) / (350 × 10⁻⁹ m)
E = 5.66 × 10⁻¹⁹ J
2. Threshold frequency:
The threshold frequency f₀ is the minimum frequency of light required to eject an electron from the metal surface. It is related to the work function φ by the equation:
φ = hf₀
where f₀ is the threshold frequency.
Substituting the given values, we get:
f₀ = φ/h
f₀ = (2.2 × 1.6 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J s)
f₀ = 5.11 × 10¹⁴ Hz
3. Number of photons:
The intensity of light is given by the equation:
I = nhf
where n is the number of photons per unit area per unit time, h is Planck’s constant, and f is the frequency of the light.
Rearranging the equation, we get:
n = I / hf
Substituting the given values, we get:
n = (1.00 W/m²) / [(6.626 × 10⁻³⁴ J s) × (3 × 10⁸ m/s) / (350 × 10⁻⁹ m)]
n = 5.70 × 10¹⁵ photons/m²s
4. Number of photoelectrons:
The number of photoelectrons emitted per unit time is given by the equation:
I₀ = nAq
where I₀ is the current due to photoelectrons, q is the charge of an electron, and A is the area of the metal surface.
The number of photoelectrons emitted per second is equal to I₀/q.
Substituting the given values, we get:
I₀ = nAq = (5.70 × 10¹⁵ photons/m²s) × (1.00 × 10⁻⁴ m²) × (0.50%) × (1.6 × 10⁻¹⁹ C/photon)
I₀ = 4.59 × 10⁻¹² A
Wavelength of ultraviolet light, λ = 350 nm = 350 × 10⁻⁹ m
Intensity of light, I = 1.00 W/m²
Work function of metal surface, φ = 2.2 eV = 2.2 × 1.6 × 10⁻¹⁹ J
Percentage of photons producing photoelectrons, n = 0.50%
Area of metal surface, A = 1.00 cm² = 1.00 × 10⁻⁴ m²
Planck’s constant, h = 6.626 × 10⁻³⁴ J s
Speed of light, c = 3 × 10⁸ m/s
To find: Number of photoelectrons emitted per second
Solution:
1. Energy of photons:
The energy of a photon is given by the equation:
E = hc/λ
where h is Planck’s constant and c is the speed of light.
Substituting the given values, we get:
E = (6.626 × 10⁻³⁴ J s) × (3 × 10⁸ m/s) / (350 × 10⁻⁹ m)
E = 5.66 × 10⁻¹⁹ J
2. Threshold frequency:
The threshold frequency f₀ is the minimum frequency of light required to eject an electron from the metal surface. It is related to the work function φ by the equation:
φ = hf₀
where f₀ is the threshold frequency.
Substituting the given values, we get:
f₀ = φ/h
f₀ = (2.2 × 1.6 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J s)
f₀ = 5.11 × 10¹⁴ Hz
3. Number of photons:
The intensity of light is given by the equation:
I = nhf
where n is the number of photons per unit area per unit time, h is Planck’s constant, and f is the frequency of the light.
Rearranging the equation, we get:
n = I / hf
Substituting the given values, we get:
n = (1.00 W/m²) / [(6.626 × 10⁻³⁴ J s) × (3 × 10⁸ m/s) / (350 × 10⁻⁹ m)]
n = 5.70 × 10¹⁵ photons/m²s
4. Number of photoelectrons:
The number of photoelectrons emitted per unit time is given by the equation:
I₀ = nAq
where I₀ is the current due to photoelectrons, q is the charge of an electron, and A is the area of the metal surface.
The number of photoelectrons emitted per second is equal to I₀/q.
Substituting the given values, we get:
I₀ = nAq = (5.70 × 10¹⁵ photons/m²s) × (1.00 × 10⁻⁴ m²) × (0.50%) × (1.6 × 10⁻¹⁹ C/photon)
I₀ = 4.59 × 10⁻¹² A
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Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer?
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Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer?.
Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer?.
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Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer?, a detailed solution for Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer? has been provided alongside types of Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Ultraviolet light of wavelength 350 nmand intensity 1.00 W/m2is directed at a metal surface of work function 2.2 eV. If 0.50% of the incident photons produce photoelectrons, the number of photo electrons emitted per second, if the metal surface lias an area of 1.00 cm2, is equal to ____ x 1011 photoelectrons/s. (upto one decimal place) (h = 6.626 x 10-34 J s, c = 3 x 108 m/s,1 eV = 1.6 x 10-19 J)Correct answer is between '8.7,8.9'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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