Answer:
The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].
Explanation:
Given that,
Mass [tex]M=6.65\times10^{-27}\ kg[/tex]
Temperature = 20.0°C
We need to calculate the root-mean square speed
Using formula of root mean square speed
[tex]v_{rms}=\sqrt{\dfrac{3kTN_{A}}{M}}[/tex]
Where, N = Avogadro number
M = Molar mass
T = Temperature
k = Boltzmann constant
Put the value into the formula
[tex]v_{rms}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times293\times6.022\times10^{23}}{4\times10^{-3}}}[/tex]
[tex]v_{rms}=1351.37\ m/s[/tex]
We need to calculate the de Broglie wavelength
Using formula of de Broglie wavelength
[tex]P=\dfrac{h}{\lambda}[/tex]
[tex]mv=\dfrac{h}{\lambda}[/tex]
[tex]\lambda=\dfrac{6.626\times10^{-34}}{6.65\times10^{-27}\times1351.37}[/tex]
[tex]\lambda=7.373\times10^{-11}\ m[/tex]
Hence, The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].
The de Broglie wavelength of the helium atoms is mathematically given as
7.373*10^{-11} m.
Question Parameter(s):
A gas of helium atoms (each of mass 6.65 × 10-27 kg) is at room temperature of 20.0°C.
Generally, the equation for the Speed is mathematically given as
[tex]v_r =\sqrt{\frac{3kTN_{A}}{M}}[/tex]
Therefore
[tex]v_{rms}=\sqrt{\frac{3*1.38*10^{-23}*293*6.022*10^{23}}{4*10^{-3}}}[/tex]
v_r=1351.37 m/s
In conclusion, de Broglie wavelength is
[tex]p=h/\lambda[/tex]
Therefore
[tex]\lambda=\frac{6.626*10^{-34}}{6.65*10^{-27}*1351.37}[/tex]
7.373*10^{-11} m.
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