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In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is?
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In one experiment a proton having initial kinetic energy of 1eV is acc...
Ratio of de-Broglie wavelengths of proton and a-particle
To find the ratio of de-Broglie wavelengths of proton and a-particle, we need to use the formula for de-Broglie wavelength:
λ = h/p
where λ is the de-Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.
Experiment 1: Proton accelerated through a potential difference of 3 V
The initial kinetic energy of the proton is 1 eV. To find the momentum of the proton, we can use the formula:
K = p²/2m
where K is the kinetic energy, p is the momentum, and m is the mass of the proton.
Rearranging the formula, we get:
p = sqrt(2mK)
Substituting the values, we get:
p = sqrt(2*1.67e-27*1.6e-19) = 3.26e-22 kg m/s
Using the formula for de-Broglie wavelength, we get:
λ_proton = h/p = 6.63e-34/3.26e-22 = 2.03e-12 m
Experiment 2: Alpha particle retarded by a potential difference of 2 V
The initial kinetic energy of the alpha particle is 20 eV. To find the momentum of the alpha particle, we can use the same formula:
K = p²/2m
Rearranging the formula, we get:
p = sqrt(2mK)
Substituting the values, we get:
p = sqrt(2*6.64e-27*20*1.6e-19) = 9.13e-22 kg m/s
Using the formula for de-Broglie wavelength, we get:
λ_alpha = h/p = 6.63e-34/9.13e-22 = 7.26e-13 m
Ratio of de-Broglie wavelengths
To find the ratio of de-Broglie wavelengths, we can divide λ_proton by λ_alpha:
λ_proton/λ_alpha = 2.03e-12/7.26e-13 = 2.79
Therefore, the ratio of de-Broglie wavelengths of proton and alpha particle is 2.79.
Explanation
The de-Broglie wavelength of a particle is inversely proportional to its momentum. In the first experiment, the proton is accelerated through a potential difference, which increases its kinetic energy and hence its momentum. In the second experiment, the alpha particle is retarded by a potential difference, which decreases its kinetic energy and hence its momentum. Since the proton has a smaller mass than the alpha particle, it gains more momentum for the same increase in kinetic energy. This results in a smaller de-Broglie wavelength for the proton compared to the alpha particle.
To find the ratio of de-Broglie wavelengths of proton and a-particle, we need to use the formula for de-Broglie wavelength:
λ = h/p
where λ is the de-Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.
Experiment 1: Proton accelerated through a potential difference of 3 V
The initial kinetic energy of the proton is 1 eV. To find the momentum of the proton, we can use the formula:
K = p²/2m
where K is the kinetic energy, p is the momentum, and m is the mass of the proton.
Rearranging the formula, we get:
p = sqrt(2mK)
Substituting the values, we get:
p = sqrt(2*1.67e-27*1.6e-19) = 3.26e-22 kg m/s
Using the formula for de-Broglie wavelength, we get:
λ_proton = h/p = 6.63e-34/3.26e-22 = 2.03e-12 m
Experiment 2: Alpha particle retarded by a potential difference of 2 V
The initial kinetic energy of the alpha particle is 20 eV. To find the momentum of the alpha particle, we can use the same formula:
K = p²/2m
Rearranging the formula, we get:
p = sqrt(2mK)
Substituting the values, we get:
p = sqrt(2*6.64e-27*20*1.6e-19) = 9.13e-22 kg m/s
Using the formula for de-Broglie wavelength, we get:
λ_alpha = h/p = 6.63e-34/9.13e-22 = 7.26e-13 m
Ratio of de-Broglie wavelengths
To find the ratio of de-Broglie wavelengths, we can divide λ_proton by λ_alpha:
λ_proton/λ_alpha = 2.03e-12/7.26e-13 = 2.79
Therefore, the ratio of de-Broglie wavelengths of proton and alpha particle is 2.79.
Explanation
The de-Broglie wavelength of a particle is inversely proportional to its momentum. In the first experiment, the proton is accelerated through a potential difference, which increases its kinetic energy and hence its momentum. In the second experiment, the alpha particle is retarded by a potential difference, which decreases its kinetic energy and hence its momentum. Since the proton has a smaller mass than the alpha particle, it gains more momentum for the same increase in kinetic energy. This results in a smaller de-Broglie wavelength for the proton compared to the alpha particle.
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In one experiment a proton having initial kinetic energy of 1eV is acc...
Abhinav ultraviolet light of all wavelength passes through hydrogen gas at room temperature in the x minus direction assume that all photons emitted due to electron transition inside the gas emerge in the Y - direction let A and B denote the light emerging from the gas in the X and Y direction respectively
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In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is?
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In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is?.
In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In one experiment a proton having initial kinetic energy of 1eV is accelerated through a potential difference of 3 V. In another experiment, an a-particle having initial kinetic energy 20 eV is retarded by a potential difference of 2 V. the ratio of de-Broglie wavelengths of proton and a-particle is?.
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