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.
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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