Chapter 28, Problem 30P

To determine

The shortest and longest wavelengths.

Answer to Problem 30P

The shortest wavelengths is $\text{0}\text{.037}\text{nm}$ .

The longest wavelengths is $\text{1}\text{nm}$ .

Explanation of Solution

Given info:

Voltage of TV picture tube $\text{=}\text{33}\text{.}\text{5}\text{kV}\text{=}\text{33}\text{.5}\times {\text{10}}^3\text{eV}$

The shortest wavelength is equals to the maximum kinetic energy of electron.

Formula used:

  $\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\text{E}}$

Where, $\text{λ}$ is a wavelength and $\text{E}$ is energy in eV.

Calculation:

The shortest wavelength X-ray has the most energy. This energy is the maximum kinetic energy of the electron in the tube:

  $\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\text{E}}\mathrm{...}\left(1\right)$

Substitute the value of E in equation $\left(1\right)$

  $\begin{array}{l}=\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left(33.5\times {10}^3\text{eV}\right)}\\ =\text{0}\text{.037}\text{nm}\end{array}$

Also, the longest wavelength of continuous spectrum will be at the limit of the X-ray region of electromagnetic spectrum, on the order of $\text{1}\text{nm}$ .

Conclusion:

The shortest wavelengths is $\text{0}\text{.037}\text{nm}$ .

The longest wavelengths is $\text{1}\text{nm}$ .