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In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?
- a)1 eV
- b)3.25 eV
- c)4 eV
- d)9.25 eV
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a photoelectric effect experiment, for radiation with frequency v0 ...
The kinetic energy of photoelectron is given as
KE = hv-W
where v → frequency o f incident photon
W → work function of the substance
h→ Planck constant
Now, hv0 = 8 eV, KE = 2eV
So, work function = (8 - 2) eV
W = 6 eV
Now, if hv0 = 8 eV
⇒ 1.25 hv0 = 10 eV
Thus, hv = 10 e V , W = 6 eV
⇒ Kinetic energy = hv - W = 10 eV - 6 eV = 4 eV
KE = hv-W
where v → frequency o f incident photon
W → work function of the substance
h→ Planck constant
Now, hv0 = 8 eV, KE = 2eV
So, work function = (8 - 2) eV
W = 6 eV
Now, if hv0 = 8 eV
⇒ 1.25 hv0 = 10 eV
Thus, hv = 10 e V , W = 6 eV
⇒ Kinetic energy = hv - W = 10 eV - 6 eV = 4 eV
Most Upvoted Answer
In a photoelectric effect experiment, for radiation with frequency v0 ...
Introduction:
The photoelectric effect refers to the phenomenon where electrons are emitted from a material surface when exposed to electromagnetic radiation. This effect can be explained by the particle-like behavior of light, where photons transfer their energy to electrons, causing them to be ejected from the material. The energy of the emitted electrons depends on the frequency of the incident radiation.
Given information:
- Frequency of the incident radiation, v0
- Energy of the incident radiation, hv0 = 8 eV
- Energy of the emitted electrons, 2 eV
- Frequency of the new radiation, 1.25 v0
Explanation:
The energy of a photon can be calculated using the equation E = hv, where E is the energy, h is Planck's constant (6.626 x 10^-34 J.s), and v is the frequency of the radiation. In this case, we are given the energy of the incident radiation and need to find the energy of the emitted electrons for a different frequency.
Step 1: Calculate the energy of the incident radiation:
Given: hv0 = 8 eV
We know that 1 eV is equal to 1.6 x 10^-19 J (1 eV = 1.6 x 10^-19 J).
Therefore, the energy of the incident radiation can be calculated as:
E0 = hv0 = 8 eV * (1.6 x 10^-19 J/eV) = 1.28 x 10^-18 J
Step 2: Calculate the energy of the new radiation:
Given: Frequency of the new radiation = 1.25 v0
The energy of the new radiation can be calculated using the equation E = hv:
E1 = h(1.25 v0)
Step 3: Calculate the energy of the emitted electrons:
Given: Energy of the emitted electrons = 2 eV
We convert this energy to joules:
E_electron = 2 eV * (1.6 x 10^-19 J/eV) = 3.2 x 10^-19 J
Step 4: Determine the energy of the emitted electrons for the new radiation:
Since energy is conserved in the photoelectric effect, the energy of the emitted electrons should be the same for both incidents of radiation. Therefore, we can equate the energy of the emitted electrons to the energy of the new radiation:
E1 = E_electron
Step 5: Substitute the values:
h(1.25 v0) = 3.2 x 10^-19 J
Step 6: Solve for the energy of the new radiation:
We know that h = 6.626 x 10^-34 J.s, so we can substitute this value and solve for the new energy:
(6.626 x 10^-34 J.s)(1.25 v0) = 3.2 x 10^-19 J
v0 = (3.2 x 10^-19 J) / (6.626 x 10^-34 J.s)(1.25)
v0 ≈ 4.854 x 10^14 Hz
Step 7: Calculate the energy of the new radiation:
E1
The photoelectric effect refers to the phenomenon where electrons are emitted from a material surface when exposed to electromagnetic radiation. This effect can be explained by the particle-like behavior of light, where photons transfer their energy to electrons, causing them to be ejected from the material. The energy of the emitted electrons depends on the frequency of the incident radiation.
Given information:
- Frequency of the incident radiation, v0
- Energy of the incident radiation, hv0 = 8 eV
- Energy of the emitted electrons, 2 eV
- Frequency of the new radiation, 1.25 v0
Explanation:
The energy of a photon can be calculated using the equation E = hv, where E is the energy, h is Planck's constant (6.626 x 10^-34 J.s), and v is the frequency of the radiation. In this case, we are given the energy of the incident radiation and need to find the energy of the emitted electrons for a different frequency.
Step 1: Calculate the energy of the incident radiation:
Given: hv0 = 8 eV
We know that 1 eV is equal to 1.6 x 10^-19 J (1 eV = 1.6 x 10^-19 J).
Therefore, the energy of the incident radiation can be calculated as:
E0 = hv0 = 8 eV * (1.6 x 10^-19 J/eV) = 1.28 x 10^-18 J
Step 2: Calculate the energy of the new radiation:
Given: Frequency of the new radiation = 1.25 v0
The energy of the new radiation can be calculated using the equation E = hv:
E1 = h(1.25 v0)
Step 3: Calculate the energy of the emitted electrons:
Given: Energy of the emitted electrons = 2 eV
We convert this energy to joules:
E_electron = 2 eV * (1.6 x 10^-19 J/eV) = 3.2 x 10^-19 J
Step 4: Determine the energy of the emitted electrons for the new radiation:
Since energy is conserved in the photoelectric effect, the energy of the emitted electrons should be the same for both incidents of radiation. Therefore, we can equate the energy of the emitted electrons to the energy of the new radiation:
E1 = E_electron
Step 5: Substitute the values:
h(1.25 v0) = 3.2 x 10^-19 J
Step 6: Solve for the energy of the new radiation:
We know that h = 6.626 x 10^-34 J.s, so we can substitute this value and solve for the new energy:
(6.626 x 10^-34 J.s)(1.25 v0) = 3.2 x 10^-19 J
v0 = (3.2 x 10^-19 J) / (6.626 x 10^-34 J.s)(1.25)
v0 ≈ 4.854 x 10^14 Hz
Step 7: Calculate the energy of the new radiation:
E1
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In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?a)1 eVb)3.25 eVc)4 eVd)9.25 eVCorrect answer is option 'C'. Can you explain this answer?
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In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?a)1 eVb)3.25 eVc)4 eVd)9.25 eVCorrect answer is option 'C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?a)1 eVb)3.25 eVc)4 eVd)9.25 eVCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?a)1 eVb)3.25 eVc)4 eVd)9.25 eVCorrect answer is option 'C'. Can you explain this answer?.
In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?a)1 eVb)3.25 eVc)4 eVd)9.25 eVCorrect answer is option 'C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?a)1 eVb)3.25 eVc)4 eVd)9.25 eVCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a photoelectric effect experiment, for radiation with frequency v0 with hv0 = 8 ev, electrons are emitted with energy 2 eV. What is the energy of the electrons emitted for incoming radiation of frequency 1.25 v0?a)1 eVb)3.25 eVc)4 eVd)9.25 eVCorrect answer is option 'C'. Can you explain this answer?.
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