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
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
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