Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

Answer to Problem 32P

The de Broglie wavelength of the electron is $5.79\times {10}^{-10}\text{m}$.

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

${\lambda }_e=\frac{h}{\sqrt{2m_e\left(KE\right)}}$

• • $h$ is the Plank’s constant
  • $m_e$ is the mass of the electron
  • $KE$ is the kinetic energy

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$ and $4.50\text{eV}$ for $KE$.

$\begin{array}{c}{\lambda }_e=\frac{6.63\times {10}^{-34}\text{J}\text{.s}}{\sqrt{2\left(9.11\times {10}^{-31}\text{kg}\right)\left(4.50\text{eV}\right)\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)}}\\ =5.79\times {10}^{-10}\text{m}\end{array}$

Conclusion:

Thus, the de Broglie wavelength of the electron is $5.79\times {10}^{-10}\text{m}$.

(b)

To determine

The de Broglie wavelength of the bowling ball.

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is $2.26\times {10}^{-25}\text{m}$.

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

${\lambda }_b=\frac{h}{\sqrt{2m_b\left(KE\right)}}$

• • $m_b$ is the mass of the bowling ball

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $6.00\text{kg}$ for $m_b$ and $4.50\text{eV}$ for $KE$.

$\begin{array}{c}{\lambda }_b=\frac{6.63\times {10}^{-34}\text{J}\text{.s}}{\sqrt{2\left(6.00\text{kg}\right)\left(4.50\text{eV}\right)\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)}}\\ =2.26\times {10}^{-25}\text{m}\end{array}$

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is $2.26\times {10}^{-25}\text{m}$.

(c)

To determine

The de Broglie wavelength of the photon.

Answer to Problem 32P

The de Broglie wavelength of the photon is $2.76\times {10}^{-7}\text{m}$.

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

${\lambda }_p=\frac{hc}{E}$

• • $c$ is the speed of light
  • $E$ is the energy of the photon

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $3.00\times {10}^8\text{m/s}$ for $c$ and $4.50\text{eV}$ for $E$.

$\begin{array}{c}{\lambda }_p=\frac{\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\left(3.00\times {10}^8\text{m/s}\right)}{\left(4.50\text{eV}\right)\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =2.76\times {10}^{-7}\text{m}\end{array}$

Conclusion:

Thus, the de Broglie wavelength of the photon is $2.76\times {10}^{-7}\text{m}$