Chapter 30, Problem 99A

(a)

To determine

To calculate:The velocity of electron.

Answer to Problem 99A

The velocity of electron is $1840\text{m/s}$ .

Explanation of Solution

Given:

  $\lambda \text{=400}\text{.0}\text{nm}$

Formula used:

De Broglie’s Wavelength is given by,

  $\lambda \text{=}\frac{h}{mv}$

Where,

  $\lambda$ : Wavelength

  $h$ : Planck’s Constant $\left(6.63\times {10}^{-34}\right)=\left(6.63e-34\right)$

  $m$ : Mass

  $v$ : Velocity

Unit conversions,

Calculation:

To find the velocity of an electron, first

  $\lambda \text{=}\frac{h}{mv}$

Also $P=mv$ ,

Rearranging the formula to find $p$ ,

  $p\text{=}\frac{h}{\lambda }$

Substitute all the values in given formula,

  $\begin{array}{l}p=\frac{6.63\times {10}^{-34}}{400\times {10}^{-9}}\\ p=1.657\times {10}^{-27}\text{kg}\cdot \text{m/s}\end{array}$

Now to calculate the velocity ,

  $v=\frac{p}{m}$

Substitute all the values in given formula,

  $\begin{array}{l}v=\frac{1.657\times {10}^{-27}\text{kg}\cdot \text{m/s}}{9.01\times {10}^{-31}\text{Kg}}\\ =\frac{1.657\times {10}^{-27}\text{kg}\cdot \text{m/s}}{9.01\times {10}^{-31}\text{kg}}\\ v=1840\text{m/s}\end{array}$

Conclusion:

The velocity of electron is $1840\text{m/s}$ .

(b)

To determine

To calculate:The energy of electron in $\text{eV}$ .

Answer to Problem 99A

The energy of electron in $\text{eV}$ is $3.10\text{eV}$ .

Explanation of Solution

Given:

  $\lambda \text{=400}\text{.0}\text{nm}$

Formula used:

To calculate the energy of electron in $\text{eV}$ is,$E=\frac{hc}{\lambda }$

Where,

  $E$ : Energy of an electron

  $h$ : Planks constant

  $c$ : Speed of light

  $\lambda$ : Wavelength

Value of $hc$ in $\text{eV}\cdot \text{nm}$

  $\text{=1240}\text{eV}\cdot \text{nm}$

Calculation:

To calculate the energy of electron in $\text{eV}$ is,$E=\frac{hc}{\lambda }$

Now, substitute all the values in the given formula,

  $E=\frac{\text{1240}\text{eV}\cdot \text{nm}}{400\text{nm}}$

  $E=3.10\text{eV}$

Conclusion:

The energy of electron in $\text{eV}$ is $3.10\text{eV}$ .