Chapter 27, Problem 56A

(a)

To determine

To Find: The velocity of an electron.

Answer to Problem 56A

  $2.19\times {10}^6\text{m/s}$

Explanation of Solution

Given data:

Kinetic energy of electron, $K.E=13.65\text{eV}$

Formula used:

  $K.E=\frac{1}{2}m{v}^2$

Here, m is the mass and v is the velocity.

Calculation:

  $\begin{array}{l}\frac{1}{2}m{v}^2=13.65\text{eV}\\ \frac{1}{2}m{v}^2=13.65\times 1.6\times {10}^{-19}\\ \frac{1}{2}m{v}^2=21.84\times {10}^{-19}\\ \frac{1}{2}\times 9.1\times {10}^{-31}\times v^2=21.85\times {10}^{-19}\\ v=\sqrt{\frac{2\times 21.85\times {10}^{-19}}{9.1\times {10}^{-31}}}\\ v=2.19\times {10}^6\text{m/s}\end{array}$

Conclusion:

Velocity of the electron is $2.19\times {10}^6\text{m/s}$ .

(b)

To determine

To Find: The de Broglie wavelength of the electron.

Answer to Problem 56A

  $3.3\stackrel{\text{o}}{\text{A}}$

Explanation of Solution

Given:

Velocity of the electron is $2.19\times {10}^6\text{m/s}$ .

Formula used:

de Broglie equation is:

  $\lambda =\frac{h}{mv}$

  $\lambda =$ de Broglie wavelength

  $h=$ Planck’s constant

m = mass

v = velocity

Calculation:

Substitute the known values and solve:

  $\lambda =\frac{h}{mv}$

  $=\frac{6.63\times {10}^{-34}}{9.1\times {10}^{-31}\times 2.19\times {10}^6}$

  $\begin{array}{l}\lambda =0.33\times {10}^{-9}\text{m}\\ \lambda =3.3\stackrel{\text{o}}{\text{A}}\end{array}$

Conclusion:

Electron’s de Broglie wavelength is $3.3\stackrel{\text{o}}{\text{A}}$ .

(c)

To determine

To Compare: The circumference of a hydrogen atom with atom’s electron de Broglie wavelength.

Explanation of Solution

Given:

Electron’s de Broglie wavelength is $3.3\stackrel{\text{o}}{\text{A}}$ .

Radius of Hydrogen atom, $r=0.519\text{nm}$

Formula used:

Circumference $=2\pi r$

Calculation:

  $\begin{array}{c}s=2\times 3.14\times 0.519\text{nm}\\ =3.26\text{nm}\end{array}$

de Broglie wavelength of electron $=0.33\text{nm}$

Ratio of circumference of hydrogen atom and de Broglie’s wavelength of electron is:

  $\frac{s}{\lambda }=\frac{3.26\text{nm}}{0.33\text{nm}}=9.88$

Conclusion:

The circumference of hydrogen atom is 9.88 times the de Broglie’s wavelength of electron.