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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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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.
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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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