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The temperature at which average de-Broglie wavelength of helium atom becomes0.5 nm is

Question

The temperature at which average de-Broglie wavelength of helium atom becomes 0.5 nm is

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Solution

The de Broglie wavelength is given by the formula:

λ = h / p

where h is Planck's constant and p is the momentum of the particle. For a gas, the momentum can be related to the temperature by the formula:

p = sqrt(2mk*T)

where m is the mass of the particle, k is Boltzmann's constant, and T is the temperature. We can substitute this into the de Broglie wavelength formula to get:

λ = h / sqrt(2mk*T)

We can rearrange this to solve for T:

T = (h^2) / (2mk*λ^2)

Given that λ = 0.5 nm = 0.5 * 10^-9 m, h = 6.626 * 10^-34 Js, m = 41.67 * 10^-27 kg (mass of helium atom), and k = 1.38 * 10^-23 J/K, we can substitute these values into the formula to find the temperature:

T = (6.626 * 10^-34)^2 / (241.67 * 10^-271.38 * 10^-23(0.5 * 10^-9)^2)

After calculating the above expression, we

This problem has been solved

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