Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

Answer to Problem 1P

The de Broglie wavelength is $1.26\times {10}^{-10}\text{m}$.

Explanation of Solution

The angular wave number from the wave function is $5.00\times {10}^{10}{\text{m}}^{-1}$.

Write the expression to calculate the wavelength.

  $\lambda =\frac{2\pi }{k}$

Here, $\lambda$ is the wavelength and k is the angular wave number.

Write the expression to calculate $5.00\times {10}^{10}{\text{m}}^{-1}$ for k in the above equation to calculate $\lambda$.

  $\begin{array}{c}\lambda =\frac{2\pi }{5.00\times {10}^{10}{\text{m}}^{-1}}\\ =1.26\times {10}^{-10}\text{m}\end{array}$

Conclusion:

Therefore, the de Broglie wavelength is $1.26\times {10}^{-10}\text{m}$.

(b)

To determine

The magnitude of momentum.

Answer to Problem 1P

The magnitude of momentum is $5.26\times {10}^{-24}\text{kg}\cdot \text{m}/\text{s}$.

Explanation of Solution

Write the expression to calculate the momentum.

  $p=\frac{h}{\lambda }$

Here, p is the momentum and h is the Planck’s constant.

Substitute $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for h and $1.26\times {10}^{-10}\text{m}$ for $\lambda$ in the above equation to calculate p.

  $\begin{array}{c}p=\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{1.26\times {10}^{-10}\text{m}}\\ =5.26\times {10}^{-24}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the magnitude of momentum is $5.26\times {10}^{-24}\text{kg}\cdot \text{m}/\text{s}$.

(c)

To determine

The kinetic energy in electron volts.

Answer to Problem 1P

The kinetic energy in electron volts is $95.0\text{eV}$.

Explanation of Solution

Write the expression to calculate the kinetic energy.

  $K=\frac{p^2}{2m}$

Here, K is the kinetic energy and m is the mass of electron.

Substitute $5.26\times {10}^{-24}\text{kg}\cdot \text{m}/\text{s}$ for p and $9.11\times {10}^{-31}\text{kg}$ for m in the above equation to calculate K.

  $\begin{array}{c}K=\frac{{\left(5.26\times {10}^{-24}\text{kg}\cdot \text{m}/\text{s}\right)}^2}{2\left(9.11\times {10}^{-31}\text{kg}\right)}\\ =1.52\times {10}^{-17}\text{J}\left(\frac{1\text{eV}}{1.602\times {10}^{-19}\text{J}}\right)\\ =95.0\text{eV}\end{array}$

Conclusion:

Therefore, the kinetic energy in electron volts is $95.0\text{eV}$.