91Ó°ÊÓ

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

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:

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

Here,${\mathrm{λ}}_{\min }$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;$1.6×{10}^{−19}\text{ C}$;$h$ is the Plank’s constant, whose value is$6.63×{10}^{−34}\text{ J}â‹\dots \text{s;}$and$c$ is the speed of light in vacuum, whose value is$3×{10}^8\text{″¾/²õ}$.

Replaceall the known values in the above equation

$\begin{array}{c}{\mathrm{λ}}_{\min }=\frac{(6.63×{10}^{−34}\text{ J}â‹\dots \text{s})(3×{10}^8\text{ }m/s)}{(1.6×{10}^{−19}\text{ C})\left(30000\text{ V}\right)}\\ =(4.14×{10}^{−11}Jâ‹\dots m/Câ‹\dots V)\left(\frac{1\text{m}}{1Jâ‹\dots m/Câ‹\dots V}\right)\\ =4.14×{10}^{−11}\text{″¾.}\end{array}$

Thus, the minimum wavelength to be expected in X-rays produced in the given tubeis $4.14×{10}^{−11}\text{″¾}$.