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Light of two different frequencies whose photons have energies of 1eV and 2.5 eV respectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons is
- a)1:5
- b)1:4
- c)1:3
- d)1:2
Correct answer is option 'D'. Can you explain this answer?
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Light of two different frequencies whose photons have energies of 1eV ...
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Light of two different frequencies whose photons have energies of 1eV ...
Concept: The maximum kinetic energy of emitted electrons is given by the equation:
KE(max) = hf - Φ
where h is Planck's constant, f is the frequency of the incident photon, and Φ is the work function of the metal. The maximum speed of the emitted electrons is proportional to the square root of their kinetic energy.
Explanation:
Given: Energy of photon 1 = 1 eV, Energy of photon 2 = 2.5 eV, Work function of metal = 0.5 eV
Let's first calculate the frequencies of the two photons using the formula:
E = hf
where E is the energy of the photon and f is its frequency.
For photon 1:
1 eV = hf
f = 1/h = 1.6 x 10^-19 Hz
For photon 2:
2.5 eV = hf
f = 2.5/h = 4 x 10^-19 Hz
Now, let's calculate the maximum kinetic energy of the emitted electrons for each frequency:
For photon 1:
KE(max) = hf - Φ = (1.6 x 10^-19 J)(1.6 x 10^-19 Hz) - (0.5 eV)(1.6 x 10^-19 J/eV) = 0.6 eV
For photon 2:
KE(max) = hf - Φ = (2.5 x 1.6 x 10^-19 J)(4 x 10^-19 Hz) - (0.5 eV)(1.6 x 10^-19 J/eV) = 1.5 eV
The ratio of the maximum speeds of the emitted electrons is given by:
(sqrt(KE(max) for photon 2) / sqrt(KE(max) for photon 1))
= sqrt(1.5 eV) / sqrt(0.6 eV)
= 1.22
Therefore, the ratio of the maximum speeds of the emitted electrons is 1:2, which is option D.
KE(max) = hf - Φ
where h is Planck's constant, f is the frequency of the incident photon, and Φ is the work function of the metal. The maximum speed of the emitted electrons is proportional to the square root of their kinetic energy.
Explanation:
Given: Energy of photon 1 = 1 eV, Energy of photon 2 = 2.5 eV, Work function of metal = 0.5 eV
Let's first calculate the frequencies of the two photons using the formula:
E = hf
where E is the energy of the photon and f is its frequency.
For photon 1:
1 eV = hf
f = 1/h = 1.6 x 10^-19 Hz
For photon 2:
2.5 eV = hf
f = 2.5/h = 4 x 10^-19 Hz
Now, let's calculate the maximum kinetic energy of the emitted electrons for each frequency:
For photon 1:
KE(max) = hf - Φ = (1.6 x 10^-19 J)(1.6 x 10^-19 Hz) - (0.5 eV)(1.6 x 10^-19 J/eV) = 0.6 eV
For photon 2:
KE(max) = hf - Φ = (2.5 x 1.6 x 10^-19 J)(4 x 10^-19 Hz) - (0.5 eV)(1.6 x 10^-19 J/eV) = 1.5 eV
The ratio of the maximum speeds of the emitted electrons is given by:
(sqrt(KE(max) for photon 2) / sqrt(KE(max) for photon 1))
= sqrt(1.5 eV) / sqrt(0.6 eV)
= 1.22
Therefore, the ratio of the maximum speeds of the emitted electrons is 1:2, which is option D.
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Light of two different frequencies whose photons have energies of 1eV and 2.5 eVrespectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons isa)1:5b)1:4c)1:3d)1:2Correct answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Light of two different frequencies whose photons have energies of 1eV and 2.5 eVrespectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons isa)1:5b)1:4c)1:3d)1:2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Light of two different frequencies whose photons have energies of 1eV and 2.5 eVrespectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons isa)1:5b)1:4c)1:3d)1:2Correct answer is option 'D'. Can you explain this answer?.
Light of two different frequencies whose photons have energies of 1eV and 2.5 eVrespectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons isa)1:5b)1:4c)1:3d)1:2Correct answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Light of two different frequencies whose photons have energies of 1eV and 2.5 eVrespectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons isa)1:5b)1:4c)1:3d)1:2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Light of two different frequencies whose photons have energies of 1eV and 2.5 eVrespectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons isa)1:5b)1:4c)1:3d)1:2Correct answer is option 'D'. Can you explain this answer?.
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