Physics Problems & Solutions: Ground State

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Quantum Mechanics - Harmonic Oscillator

If v is frequency and h is Planck’s constant, the ground state energy of a one-dimensional quantum mechanical harmonic oscillator

A. 0
B. 1/3 hv
C. 1/2 hv
D. hv
E. 3/2 hv

(GR9277 #56)

Solution:

The energy of a 1-D quantum HO: E_n = (n + ½)hv

The ground state energy, n = 0 → E₀ = ½ hv

Answer: C

Nuclear & Particle Physics - Electron Configuration

The ground state configuration of a neutral sodium atom (Z = 11) is

A. 1s² 2s² 2p⁵ 3s²
B. 1s² 2s³ 2p⁶
C. 1s² 2s² 2p⁶ 3s
D. 1s² 2s² 2p⁶ 3p
E. 1s² 2s² 2p⁵

(GR9277 #58)

Solution:

Z = 11 → 1s²2s²2p⁶3s¹

Answer: C

Note:

Electron configuration:

	l = 0	l = 1	l = 2	l = 3	# of electrons
n = 1	1s²				2
n = 2	2s²	2p⁶			8
n = 3	3s²	3p⁶	3d¹⁰		18
n = 4	4s²	4p⁶	4d¹⁰	4f¹⁴	32

Quantum Mechanics - Bohr Radius

The solution to the Schrodinger equation for the ground state of hydrogen is

$\psi_0=\frac{1}{\sqrt{\pi a^3_0}}e^{-r/a_0}$

where a₀ is the Bohr radius and r is the distance from the origin. Which of the following is the most probable value for r?

A. 0
B. a₀ / 2
C. a₀
D. 2a₀
E. ∞

(GR0177 #93)

Solution:

Probability

$\begin{aligned} P(r)dr &=|\psi_0|^2dV\\&=|\psi_0|^2d \left(\frac{4}{3}\pi r^3 \right)\\&=|\psi_0|^2 4\pi r^2 dr\\&=\frac{4 r^2}{a^3_0} e^{-2r/a_0}dr \end{aligned}$

The most probable value of r corresponds to the peak of the plot of P(r) versus r.
The slope of the curve at this point is zero.

$\frac{dP(r)}{dr}=0$

$\begin{aligned} \frac{dP(r)}{dr} &=\frac{d}{dr}\frac{4 r^2}{a^3_0} e^{-2r/a_0} \\&=\frac{4}{a^3_0}\frac{d}{dr}r^2e^{-2r/a_0}\\&=\frac{4}{a^3_0}\left(2re^{-2r/a_0}+ \frac{-2}{a_0}e^{-2r/a_0}r^2\right)\\&=\frac{8 r}{a^3_0}e^{-2r/a_0}\left(1-\frac{r}{a_0} \right ) \end{aligned}$

For $\frac{dP(r)}{dr}=0$

$1-\frac{r}{a_0}=0\rightarrow r = a_0$

r = a₀

Answer: C