Chapter 27, Problem 82GP

To determine

The velocity of electrons.

Answer to Problem 82GP

  $1.2\times {10}^{3}m/s$ .

Explanation of Solution

Given:

Angle, $\text{θ}=3.5°$

Split distance, $\text{d}=0.010\times {10}^{-3}\text{m}$

Formula used:

It is known that: $d\sin \theta =m\lambda \&mv=\frac{h}{\lambda }$

Where,

  $h$ is Planck’s constant.

  $v$ is velocity.

  $m$ is mass.

  $\theta$ is an angle.

  $\text{λ}$ is wavelength.

Calculation:

Finding the wavelength of the light, $d\sin \theta =m\lambda$

  $\left(0.010\times {10}^{-3}m\right)\sin 3.5°=\left(1\right)\lambda$

  $\therefore \lambda =6.10\times {10}^{-7}m$

The velocity of the electron that has the same wavelength, $mv=\frac{h}{\lambda }$

  $\left(9.11\times {10}^{-31}kg\right)v=\frac{6.63\times {10}^{-34}J\cdot s}{6.10\times {10}^{-7}m}$

  $v=\frac{6.63\times {10}^{-34}J\cdot s}{\left(9.11\times {10}^{-31}kg\right)6.10\times {10}^{-7}m}$

  $v=1.2\times {10}^{3}m/s$

Conclusion: The velocity of the electrons will be $1.2\times {10}^{3}m/s$ .