Chapter 28, Problem 50P

(a)

To determine

The quantum number of the marble.

Answer to Problem 50P

The quantum number of the marble is $6\times {10}^{28}$.

Explanation of Solution

The mass of the marble is $10\text{ g}$, its speed is $2{\text{cms}}^{-1}$ and the length of the box is $10\text{ cm}$.

Write the expression for momentum of a particle in a box.

$p_n=\frac{nh}{2L}$                                                            (I)

Here, $p_n$ is the momentum of the marble, $n$ is the quantum number, $h$ is the Planck’s constant and $L$ is the length of the box.

Rearranging (I) for $n$

$n=\frac{2p_{n}L}{h}$                                                            (II)

Write the classical expression for momentum.

$p_n=mv$                                                           (III)

Here, $m$ is the mass of the marble and $v$ is its speed.

Substituting (III) in (II)

$n=\frac{2mvL}{h}$                                                            (IV)

Substituting $10\text{ g}$ for $m$, $2{\text{cms}}^{-1}$ for $v$, $6.626\times {10}^{-34}\text{ Js}$ for $h$ and $10\text{ cm}$ for $L$ in (IV) to find $n$

$\begin{array}{c}n=\frac{2\times 10\text{ g}\times 2{\text{cms}}^{-1}\times 6.626\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{10\text{ cm}}\\ =\frac{2\times 10\times {10}^{-3}\text{ kg}\times 2{\text{cms}}^{-1}\times 6.626\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{10\text{ cm}}\\ =6\times {10}^{28}\end{array}$

Thus, the quantum number of the marble is $6\times {10}^{28}$.

(b)

To determine

The reason why the quantization of the marble isn’t observable.

Answer to Problem 50P

Because of the small energy difference between two quantized levels of the marble, the quantization isn’t observable.

Explanation of Solution

Write the expression for energy of the quantized level.

$E_n=\frac{n^2{h}^2}{8m{L}^2}$                                                            (V)

Here, $E_n$ is the energy of the level $n$

Substituting $n+1$ for $n$ in (I)

$E_{n+1}=\frac{{\left(n+1\right)}^2{h}^2}{8m{L}^2}$                                                            (VI)

Here, $E_{n+1}$ is the energy of the level $n+1$

Subtracting (I) from (II)

$\begin{array}{c}\Delta E=\frac{{\left(n+1\right)}^2{h}^2}{8m{L}^2}-\frac{n^2{h}^2}{8m{L}^2}\\ \Delta E=\frac{h^2}{8m{L}^2}\left[{\left(n+1\right)}^2-n^2\right]\end{array}$

Here, $\Delta E$ is the energy difference.

Thus, the energy difference is

$\Delta E=\frac{h^2}{8m{L}^2}\left(2n+1\right)$                                                           (VII)

Substituting $10\text{ g}$ for $m$, $2{\text{cms}}^{-1}$ for $v$, $6.626\times {10}^{-34}\text{ Js}$ for $h$, $10\text{ cm}$ for $L$, $6\times {10}^{28}$ for $n$ in (VII) to find $\Delta E$

 $\begin{array}{c}\Delta E=\frac{\left(2\left(6\times {10}^{28}\right)+1\right)\times {\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\times 10\text{ g}\times {\left(10\text{ cm}\right)}^2}\\ =\frac{\left(12\times {10}^{28}+1\right)\times {\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\times 10\times {10}^{-3}\text{ g}\times {\left(10\times {10}^{-2}\text{ cm}\right)}^2}\\ =7\times {19}^{-35}\text{ J}\end{array}$

The energy difference is $7\times {19}^{-35}\text{ J}$ which is very small. Thus, they cannot be observed.