Chapter 27, Problem 69A

(a)

To determine

To Find: Velocity of the electron.

Answer to Problem 69A

Velocity of electron $=1\cdot 82\times {10}^3\text{ m/s}$

Explanation of Solution

Given:

de Broglie wavelength of electron $=400\text{ nm}$

  $=4\times {10}^{-7}\text{ m}$

Formula used:

de Broglie wavelength $=\frac{h}{mv}$

Where, h = Planck’s constant

m = mass of electrons $=9\cdot 1\times {10}^{-31}\text{ kg}$

v = velocity of electron

Calculation:

  $\begin{array}{l}\text{4}\times {10}^{-7}=\frac{6\cdot 625\times {10}^{-34}}{\left(9\cdot 1\times {10}^{-31}\right)\left(v\right)}\\ v=1\cdot 82\times {10}^3\text{ m/s}\end{array}$

Conclusion:

Velocity of the electron having de Broglie $400\text{ nm}$ is $v=1\cdot 82\times {10}^3\text{m/s}$ .

(b)

To determine

To Find: Energy of the electron if its de Broglie wave length is $400\text{ nm}$ .

Answer to Problem 69A

  $3.1\text{eV}$

Explanation of Solution

Given:

de Broglie wavelength of electron $=400\text{ nm}$

  $=4\times {10}^{-7}\text{ m}$

Formula used:

Energy of electron, $\frac{hc}{\lambda }$

Here, h is the Planck’s constant, c is the speed of light and $\lambda$ is the wavelength of light.

Calculation:

  $E=\frac{hc}{\lambda }$

  $\begin{array}{l}E=\frac{6\cdot 625\times {10}^{-34}\times 3\times {10}^8}{4\times {10}^{-7}}\\ E=4\cdot 97\times {10}^{-19}\text{J}\\ E=\frac{4\cdot 97\times {10}^{-19}\text{J}}{1.6\times {10}^{-19}\text{J/eV}}=3.1\text{eV}\end{array}$

Conclusion:

The electron has energy equal to $3.1\text{eV}$ .