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The maximum wavelength of photon to break a cooper pair in Lead which has energy gap of 2.73 meV is_____________x 105 nm. (Round off to three decimal places)
Correct answer is between ''. Can you explain this answer?
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Calculation of maximum wavelength of photon to break a Cooper pair in Lead
- The energy required to break a Cooper pair is twice the energy gap of the material.
- Therefore, in this case, the energy required to break a Cooper pair in Lead is 2 × 2.73 meV = 5.46 meV.
- The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
- To break a Cooper pair, the energy of the photon must be equal to or greater than the energy required to break the pair.
- Therefore, we can write hc/λ = 5.46 meV.
- Solving for λ, we get λ = hc/5.46 meV.
- Plugging in the values of h and c and converting meV to Joules, we get λ = 2.42 × 10^5 nm or 2.42 × 10^2 μm.
- Rounding off to three decimal places, we get the answer as 242.000 x 10^3 nm or 242 μm.
The maximum wavelength of photon to break a Cooper pair in Lead with an energy gap of 2.73 meV is 242.000 x 10^3 nm.
Energy required to break Cooper pair
- The energy required to break a Cooper pair is twice the energy gap of the material.
- Therefore, in this case, the energy required to break a Cooper pair in Lead is 2 × 2.73 meV = 5.46 meV.
Calculation of maximum wavelength
- The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
- To break a Cooper pair, the energy of the photon must be equal to or greater than the energy required to break the pair.
- Therefore, we can write hc/λ = 5.46 meV.
- Solving for λ, we get λ = hc/5.46 meV.
- Plugging in the values of h and c and converting meV to Joules, we get λ = 2.42 × 10^5 nm or 2.42 × 10^2 μm.
- Rounding off to three decimal places, we get the answer as 242.000 x 10^3 nm or 242 μm.
Final Answer
The maximum wavelength of photon to break a Cooper pair in Lead with an energy gap of 2.73 meV is 242.000 x 10^3 nm.
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The maximum wavelength of photon to break a cooper pair in Lead which has energy gap of 2.73 meV is_____________x 105 nm. (Round off to three decimal places)Correct answer is between ''. Can you explain this answer?
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