a) Calculate the average de Broglie wavelength of a nitrogen molecule at room temperature, given that the mass of nitrogen molecule is 4.65 xx10^(-26)kg4.65 \times 10^{-26} \mathrm{~kg}.
Answer:
To calculate the average de Broglie wavelength of a nitrogen molecule at room temperature, we can use the formula for the de Broglie wavelength:
lambda=(h)/(p)\lambda = \frac{h}{p}
where hh is the Planck constant (6.626 xx10^(-34)” m”^(2)”kg s”^(-1)6.626 \times 10^{-34} \text{ m}^2 \text{kg s}^{-1}) and pp is the momentum of the particle.
At room temperature, the average kinetic energy of the nitrogen molecule is given by:
(:E_(k):)=(3)/(2)k_(B)T\langle E_k \rangle = \frac{3}{2} k_B T
where k_(B)k_B is the Boltzmann constant (1.38 xx10^(-23)” J K”^(-1)1.38 \times 10^{-23} \text{ J K}^{-1}) and TT is the temperature (room temperature is approximately 298 K).
The average kinetic energy can also be expressed in terms of the average momentum ((:p:)\langle p \rangle) of the nitrogen molecule:
Now, we can substitute the values for hh, mm, k_(B)k_B, and TT to calculate the average de Broglie wavelength of a nitrogen molecule at room temperature:
So, the average de Broglie wavelength of a nitrogen molecule at room temperature is approximately 1.97 xx10^(-11)” m”1.97 \times 10^{-11} \text{ m} or 19.7″ pm”19.7 \text{ pm}.
b) A particle of mass mm is constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance L_(0)L_0. Using the uncertainty principle, determine the zero-point energy of the particle.
Answer:
The zero-point energy of a particle is the lowest possible energy it can have, which is a consequence of the Heisenberg uncertainty principle. For a particle constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance L_(0)L_0, the uncertainty in position (Delta x\Delta x) can be approximated as the width of the region, L_(0)L_0.
According to the Heisenberg uncertainty principle, the product of the uncertainties in position and momentum is equal to or greater than ℏ//2\hbar/2, where ℏ\hbar is the reduced Planck constant:
Delta x*Delta p >= (ℏ)/(2)\Delta x \cdot \Delta p \geq \frac{\hbar}{2}
Solving for the uncertainty in momentum (Delta p\Delta p):
Delta p >= (ℏ)/(2Delta x)=(ℏ)/(2L_(0))\Delta p \geq \frac{\hbar}{2\Delta x} = \frac{\hbar}{2L_0}
The kinetic energy (KK) of the particle is related to its momentum (pp) by:
K=(p^(2))/(2m)K = \frac{p^2}{2m}
Using the uncertainty in momentum as an estimate for the momentum of the particle, the zero-point energy (E_(0)E_0) can be approximated as the minimum kinetic energy:
Therefore, the zero-point energy of the particle constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance L_(0)L_0 is given by:
This expression provides an estimate of the minimum energy that the particle can have due to the constraints imposed by the uncertainty principle and the spatial confinement between the potential barriers.