The given wavefunction is $\Phi(x) = x(a - x)$, which can be expanded as $\Phi(x) = ax - x^2$. We need to evaluate the truthfulness of the provided statements about this function.
A stationary state in quantum mechanics corresponds to a wavefunction that is an eigenfunction of the Hamiltonian operator ($H\Phi = E\Phi$). Such states have probability densities ($|\Psi|^2$) that are independent of time. The form $\Phi(x) = x(a-x)$ resembles the ground state wavefunction of a particle confined to an infinite potential well between $x=0$ and $x=a$. The ground state of this system is indeed a stationary state. Thus, the statement that it represents a stationary state is plausible in typical quantum mechanics contexts.
To determine if $\Phi(x)$ is an odd function, we check the condition $\Phi(-x) = -\Phi(x)$.
Comparing $\Phi(-x)$ and $-\Phi(x)$, we see that $-ax - x^2 \neq -ax + x^2$ (unless $x=0$). Therefore, $\Phi(x)$ is not an odd function.
Wavefunctions for a particle on a circular ring typically depend on the angular coordinate $\phi$, often in the form of $e^{im\phi}$. The given function $\Phi(x) = x(a - x)$ is dependent solely on the Cartesian coordinate $x$. It does not possess the characteristics required for describing a particle moving on a circular path, nor does the parameter $a$ represent the radius in this context.
The one-dimensional momentum operator is given by $\hat{p} = -i\hbar \frac{d}{dx}$. For $\Phi(x)$ to be an eigenfunction of $\hat{p}$, the action of $\hat{p}$ on $\Phi(x)$ must yield a constant multiple (the eigenvalue, $p$) of $\Phi(x)$ itself, i.e., $\hat{p}\Phi(x) = p\Phi(x)$.
Applying the momentum operator:
$ \hat{p}\Phi(x) = -i\hbar \frac{d}{dx} [x(a - x)] $ $ \hat{p}\Phi(x) = -i\hbar \frac{d}{dx} [ax - x^2] $Performing the differentiation:
$ \hat{p}\Phi(x) = -i\hbar (a - 2x) $The result, $-i\hbar (a - 2x)$, is not directly proportional to the original wavefunction $\Phi(x) = ax - x^2$. Therefore, $\Phi(x)$ is not an eigenfunction of the momentum operator.
Based on the analysis:
Thus, the statement "It represents a stationary state" is the correct characterization.
The wave function of a particle in a cubic box (of side L) is given by
$\psi(x, y, z) = \sqrt{32/L^3} \sin \frac{\pi x}{L} \cos \frac{\pi x}{L} \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L}$.
The ratio of the energy of the state corresponding to the above wave function to the ground state energy is ________.
(rounded off to the nearest integer)
The $\pi$ electrons in benzene can be modelled as particles in a ring that follow Pauli's exclusion principle. Given that the radius of benzene is 1.4 Å, the longest wavelength of light that is absorbed during an electronic transition in benzene is ____________ nm. (Up to one decimal place. Use $m_e =9.1\times10^{-31} \text{ kg}$, $h=6.6\times10^{-34} \text{ Js}$, $c=3.0\times10^8 \text{ m s}^{-1}$)