A television picture tube accelerates electrons through a potential difference of $\mathbf{30000}\mathbf{V}$. Find the minimum wavelength to be expected in X-rays produced in this tube. (Picture tubes incorporate shielding to control X-ray emission).

It is mentioned that the potential difference of a television picture tube is, $V=30000\text{\hspace{0.17em}V}$.

Whenever a specifically charged particle moves between a particular potential difference (electric) in a tube, then the generation of X-rays take place. The wavelengths of anX-ray can be obtained with the help of the relation of energy of a photon.

The relation of theminimum wavelength to be expected in X-rays, produced in the tubeis expressed as:

$\begin{array}{c}{q}_{e}V=\frac{hc}{{\lambda }_{\min }}\\ {\lambda }_{\min }=\frac{hc}{q_{e}V}\end{array}$

Here,${\lambda }_{\min }$is theminimum wavelength to be expected in an X-ray produced in the tube;$q_e$is the charge on an electron, whose value is;$1.6\times {10}^{-19}\text{\hspace{0.17em}C}$;$h$ is the Plank’s constant, whose value is$6.63\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s;}$and$c$ is the speed of light in vacuum, whose value is$3\times {10}^8\text{\hspace{0.17em}m/s}$.

Replaceall the known values in the above equation

$\begin{array}{c}{\lambda }_{\min }=\frac{(6.63\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s})(3\times {10}^{8}m/s)}{(1.6\times {10}^{-19}\text{\hspace{0.17em}C})\left(30000\text{\hspace{0.17em}V}\right)}\\ =(4.14\times {10}^{-11}J\cdot m/C\cdot V)\left(\frac{1\text{m}}{1J\cdot m/C\cdot V}\right)\\ =4.14\times {10}^{-11}\text{\hspace{0.17em}m.}\end{array}$

Thus, the minimum wavelength to be expected in X-rays produced in the given tubeis $4.14\times {10}^{-11}\text{\hspace{0.17em}m}$.