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Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?
- a)The velocity of photoelectrons increases with increase in intensity of radiation
- b)Photo-emission of electrons takes place
- c)The threshold frequency of the metal is 1.21 × 1015 sec–1
- d)The velocity of the photo-electrons is 4.48 × 106 m/sec
Correct answer is option 'B,C,D'. Can you explain this answer?
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Radiation of wavelength 200 Å falls on a platinum surface. If the wor...
According to emission photoelectric effect
E of incident radiation = hc/λ
= 9.94 × 10–18 J
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Radiation of wavelength 200 Å falls on a platinum surface. If the wor...
Explanation:
Photoelectric Effect:
The photoelectric effect refers to the emission of electrons from a material when light of certain frequencies (or wavelengths) shines on the surface of the material.
Work Function:
The work function of a metal is the minimum energy required to remove an electron from the surface of the metal. It is often denoted by the symbol φ and is measured in electron volts (eV).
Threshold Frequency:
The threshold frequency of a metal is the minimum frequency (or maximum wavelength) of light required to cause photoemission from the metal surface. It is related to the work function of the metal by the equation:
Threshold frequency (ν) = Work function (φ) / Planck's constant (h)
Given Information:
- Radiation of wavelength 200 Å falls on a platinum surface.
- Work function of the metal is 5 eV.
Analysis:
To determine the correct results about the experiment, let's analyze each option:
a) The velocity of photoelectrons increases with increase in intensity of radiation:
The velocity of photoelectrons depends on the energy of the incident photons, which is determined by the frequency (or wavelength) of the radiation. The intensity of radiation refers to the number of photons per unit area per unit time. However, the intensity does not affect the energy of individual photons. Therefore, the velocity of photoelectrons does not increase with an increase in the intensity of radiation. This option is incorrect.
b) Photo-emission of electrons takes place:
The photoelectric effect occurs when light of sufficient frequency (or energy) falls on the surface of a material. In this case, radiation of wavelength 200 Å is incident on the platinum surface. Since the wavelength is known, we can calculate the frequency of the radiation using the equation:
Speed of light (c) = Frequency (ν) x Wavelength (λ)
ν = c / λ = (3 x 10^8 m/s) / (200 x 10^-10 m) = 1.5 x 10^15 Hz
Since the frequency is greater than the threshold frequency of the metal, photoemission of electrons will take place. This option is correct.
c) The threshold frequency of the metal is 1.21 × 10^15 sec–1:
The threshold frequency can be calculated using the equation mentioned earlier:
Threshold frequency (ν) = Work function (φ) / Planck's constant (h)
Threshold frequency (ν) = 5 eV / (4.14 x 10^-15 eV s) ≈ 1.21 x 10^15 Hz
Therefore, the threshold frequency of the metal is approximately 1.21 × 10^15 Hz. This option is correct.
d) The velocity of the photo-electrons is 4.48 × 10^6 m/sec:
The velocity of photoelectrons can be calculated using the equation:
Kinetic energy (KE) = 1/2 mv^2
The energy of a photon can be calculated using the equation:
Energy of photon (E) = Planck's constant (h) x Frequency (ν)
The energy of a photon is equal to the work function of the metal, so:
Energy of photon (E) = Work function (φ
Photoelectric Effect:
The photoelectric effect refers to the emission of electrons from a material when light of certain frequencies (or wavelengths) shines on the surface of the material.
Work Function:
The work function of a metal is the minimum energy required to remove an electron from the surface of the metal. It is often denoted by the symbol φ and is measured in electron volts (eV).
Threshold Frequency:
The threshold frequency of a metal is the minimum frequency (or maximum wavelength) of light required to cause photoemission from the metal surface. It is related to the work function of the metal by the equation:
Threshold frequency (ν) = Work function (φ) / Planck's constant (h)
Given Information:
- Radiation of wavelength 200 Å falls on a platinum surface.
- Work function of the metal is 5 eV.
Analysis:
To determine the correct results about the experiment, let's analyze each option:
a) The velocity of photoelectrons increases with increase in intensity of radiation:
The velocity of photoelectrons depends on the energy of the incident photons, which is determined by the frequency (or wavelength) of the radiation. The intensity of radiation refers to the number of photons per unit area per unit time. However, the intensity does not affect the energy of individual photons. Therefore, the velocity of photoelectrons does not increase with an increase in the intensity of radiation. This option is incorrect.
b) Photo-emission of electrons takes place:
The photoelectric effect occurs when light of sufficient frequency (or energy) falls on the surface of a material. In this case, radiation of wavelength 200 Å is incident on the platinum surface. Since the wavelength is known, we can calculate the frequency of the radiation using the equation:
Speed of light (c) = Frequency (ν) x Wavelength (λ)
ν = c / λ = (3 x 10^8 m/s) / (200 x 10^-10 m) = 1.5 x 10^15 Hz
Since the frequency is greater than the threshold frequency of the metal, photoemission of electrons will take place. This option is correct.
c) The threshold frequency of the metal is 1.21 × 10^15 sec–1:
The threshold frequency can be calculated using the equation mentioned earlier:
Threshold frequency (ν) = Work function (φ) / Planck's constant (h)
Threshold frequency (ν) = 5 eV / (4.14 x 10^-15 eV s) ≈ 1.21 x 10^15 Hz
Therefore, the threshold frequency of the metal is approximately 1.21 × 10^15 Hz. This option is correct.
d) The velocity of the photo-electrons is 4.48 × 10^6 m/sec:
The velocity of photoelectrons can be calculated using the equation:
Kinetic energy (KE) = 1/2 mv^2
The energy of a photon can be calculated using the equation:
Energy of photon (E) = Planck's constant (h) x Frequency (ν)
The energy of a photon is equal to the work function of the metal, so:
Energy of photon (E) = Work function (φ
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Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer?
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Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. 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 Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer?.
Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. 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 Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer?.
Solutions for Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer?, a detailed solution for Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer? has been provided alongside types of Radiation of wavelength 200 Å falls on a platinum surface. If the work function of the metal is 5 eV.Which of the following results are correct about experiment ?a)The velocity of photoelectrons increases with increase in intensity of radiationb)Photo-emission of electrons takes placec)The threshold frequency of the metal is 1.21 × 1015 sec–1d)The velocity of the photo-electrons is 4.48 × 106 m/secCorrect answer is option 'B,C,D'. Can you explain this answer? theory, EduRev gives you an
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