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In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of -1.2 V. Then
- a)The energy of the incident photon is 4.7 eV
- b)The energy of the incident photon is 2.3 eV
- c)If higher frequency photons be used, the photoelectric current will rise
- d)When the energy of photons is 3.5 eV, the photoelectric current will be maximum
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
In photoelectric effect, the work function of a metal is 3.5 eV. The e...
Explanation:
Photoelectric effect is the phenomenon of emission of electrons by a metal when illuminated by electromagnetic radiation.
Work function: It is the minimum amount of energy required to remove an electron from the metal.
Stopping potential: It is the minimum negative potential required to stop the emitted electrons from reaching the anode.
Now, according to the given information:
Work function = 3.5 eV
Stopping potential = -1.2 V
We need to find the energy of the incident photon.
Formula:
Energy of incident photon = work function + kinetic energy of the emitted electron
Kinetic energy of the emitted electron = eV
where e is the charge of an electron and V is the stopping potential.
Substituting the given values in the formula:
Energy of incident photon = 3.5 eV + e(-1.2 V)
= 3.5 eV + (-1.2 eV)
= 2.3 eV
Therefore, the energy of the incident photon is 2.3 eV.
Option B is incorrect.
Now, let's consider the other options:
c) If higher frequency photons be used, the photoelectric current will rise
This statement is true. If higher frequency photons are used, the energy of the incident photons will increase. This will lead to the ejection of more energetic electrons, which will increase the photoelectric current.
d) When the energy of photons is 3.5 eV, the photoelectric current will be maximum
This statement is incorrect. The photoelectric current is not directly proportional to the energy of the incident photon. It depends on the number of photons incident per unit time and the number of electrons emitted per unit time.
Therefore, the correct option is A.
Photoelectric effect is the phenomenon of emission of electrons by a metal when illuminated by electromagnetic radiation.
Work function: It is the minimum amount of energy required to remove an electron from the metal.
Stopping potential: It is the minimum negative potential required to stop the emitted electrons from reaching the anode.
Now, according to the given information:
Work function = 3.5 eV
Stopping potential = -1.2 V
We need to find the energy of the incident photon.
Formula:
Energy of incident photon = work function + kinetic energy of the emitted electron
Kinetic energy of the emitted electron = eV
where e is the charge of an electron and V is the stopping potential.
Substituting the given values in the formula:
Energy of incident photon = 3.5 eV + e(-1.2 V)
= 3.5 eV + (-1.2 eV)
= 2.3 eV
Therefore, the energy of the incident photon is 2.3 eV.
Option B is incorrect.
Now, let's consider the other options:
c) If higher frequency photons be used, the photoelectric current will rise
This statement is true. If higher frequency photons are used, the energy of the incident photons will increase. This will lead to the ejection of more energetic electrons, which will increase the photoelectric current.
d) When the energy of photons is 3.5 eV, the photoelectric current will be maximum
This statement is incorrect. The photoelectric current is not directly proportional to the energy of the incident photon. It depends on the number of photons incident per unit time and the number of electrons emitted per unit time.
Therefore, the correct option is A.
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Community Answer
In photoelectric effect, the work function of a metal is 3.5 eV. The e...
Explanation:
The photoelectric effect is the phenomenon of emission of electrons from the surface of a metal when light falls on it. The energy of the incident photons determines the kinetic energy of the emitted electrons. The work function of a metal is the minimum energy required to remove an electron from its surface.
Given: Work function (W) = 3.5 eV, Stopping potential (V) = -1.2 V
To find: Energy of the incident photon (E)
Formula: Energy of photon (E) = hν = hc/λ = (1240 eV nm)/λ
where h = Planck's constant = 6.626 x 10^-34 J s
c = speed of light = 3 x 10^8 m/s
λ = wavelength of light
ν = frequency of light
Solution:
1. To find the energy of the incident photon, we need to first calculate the kinetic energy (K) of the emitted electron using the stopping potential.
K = eV = (1.6 x 10^-19 C) x (-1.2 V) = 1.92 x 10^-19 J
2. The energy of the incident photon is given by the equation:
E = K + W
where W is the work function of the metal.
E = 1.92 x 10^-19 J + 3.5 eV = (1.92 x 10^-19 J) + (3.5 x 1.6 x 10^-19 J/eV) = 4.7 eV
Therefore, the energy of the incident photon is 4.7 eV.
Answer: Option (A) The energy of the incident photon is 4.7 eV.
Additional information:
c) If higher frequency photons be used, the photoelectric current will rise.
The photoelectric current is directly proportional to the intensity of the incident light, which is determined by the frequency of the photons. Higher frequency photons have higher energy and can eject electrons with higher kinetic energy, resulting in a higher photoelectric current.
d) When the energy of photons is 3.5 eV, the photoelectric current will be maximum.
This statement is not necessarily true. The photoelectric current depends on the intensity of the incident light, which is determined by the frequency (or wavelength) of the photons. The maximum photoelectric current is obtained when the incident light has a high intensity (i.e., a large number of photons per unit area per unit time) and a frequency that corresponds to the work function of the metal or higher.
The photoelectric effect is the phenomenon of emission of electrons from the surface of a metal when light falls on it. The energy of the incident photons determines the kinetic energy of the emitted electrons. The work function of a metal is the minimum energy required to remove an electron from its surface.
Given: Work function (W) = 3.5 eV, Stopping potential (V) = -1.2 V
To find: Energy of the incident photon (E)
Formula: Energy of photon (E) = hν = hc/λ = (1240 eV nm)/λ
where h = Planck's constant = 6.626 x 10^-34 J s
c = speed of light = 3 x 10^8 m/s
λ = wavelength of light
ν = frequency of light
Solution:
1. To find the energy of the incident photon, we need to first calculate the kinetic energy (K) of the emitted electron using the stopping potential.
K = eV = (1.6 x 10^-19 C) x (-1.2 V) = 1.92 x 10^-19 J
2. The energy of the incident photon is given by the equation:
E = K + W
where W is the work function of the metal.
E = 1.92 x 10^-19 J + 3.5 eV = (1.92 x 10^-19 J) + (3.5 x 1.6 x 10^-19 J/eV) = 4.7 eV
Therefore, the energy of the incident photon is 4.7 eV.
Answer: Option (A) The energy of the incident photon is 4.7 eV.
Additional information:
c) If higher frequency photons be used, the photoelectric current will rise.
The photoelectric current is directly proportional to the intensity of the incident light, which is determined by the frequency of the photons. Higher frequency photons have higher energy and can eject electrons with higher kinetic energy, resulting in a higher photoelectric current.
d) When the energy of photons is 3.5 eV, the photoelectric current will be maximum.
This statement is not necessarily true. The photoelectric current depends on the intensity of the incident light, which is determined by the frequency (or wavelength) of the photons. The maximum photoelectric current is obtained when the incident light has a high intensity (i.e., a large number of photons per unit area per unit time) and a frequency that corresponds to the work function of the metal or higher.
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In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of -1.2 V. Thena)The energy of the incident photon is 4.7 eVb)The energy of the incident photon is 2.3 eVc)If higher frequency photons be used, the photoelectric current will rised)When the energy of photons is 3.5 eV, the photoelectric current will be maximumCorrect answer is option 'A'. Can you explain this answer?
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Solutions for In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of -1.2 V. Thena)The energy of the incident photon is 4.7 eVb)The energy of the incident photon is 2.3 eVc)If higher frequency photons be used, the photoelectric current will rised)When the energy of photons is 3.5 eV, the photoelectric current will be maximumCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of -1.2 V. Thena)The energy of the incident photon is 4.7 eVb)The energy of the incident photon is 2.3 eVc)If higher frequency photons be used, the photoelectric current will rised)When the energy of photons is 3.5 eV, the photoelectric current will be maximumCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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