Question 14 Marks
The electron beam in a colour $TV$ is accelerated through $32kV$ and then strikes the screen. What is the wavelength of the most energetic $X-$ray photon$?$
Answer
View full question & answer→$V = 32KV = 32 \times 10^3V$
When accelerated through $32KV$
$E = 32 \times 10^3eV$
$\lambda=\frac{\text{hc}}{\text{E}}=\frac{1242}{32\times10^3}$
$=38.8\times10^{-3}\text{nm}=38.8\text{pm}$
When accelerated through $32KV$
$E = 32 \times 10^3eV$
$\lambda=\frac{\text{hc}}{\text{E}}=\frac{1242}{32\times10^3}$
$=38.8\times10^{-3}\text{nm}=38.8\text{pm}$