Question
Find the
(a) Maximum frequency, and
(b) Minimum wavelength of X-rays produced by 30 kV electrons.
(a) Maximum frequency, and
(b) Minimum wavelength of X-rays produced by 30 kV electrons.
✓
Answer
(a) Given :
$\begin{aligned} V =30 kV =30 \times 1000 & =30,000 V \\ & =3 \times 10^4 V\end{aligned}$
We know that : $E =h \nu= eV$
$\therefore$ From $\quad v_{\max }=\frac{ eV }{ h }$
Where $\quad h =6.63 \times 10^{-34} Js$
$e =1.6 \times 10^{-19} C$
$=7.24 \times 10^{18} Hz$
(b) Let minimum wavelength of X-rays $=\lambda_{\min }=$ ?
$\therefore \quad \lambda_{\min }=\frac{c}{v_{\max }}$
Substituting values
$\lambda_{\min }=\frac{3 \times 10^8}{7.24 \times 10^{18}}$
$=4.14 \times 10^{-11} m$
$=4.14 \times 10^{-11} \times 10^{10}$Å
$=4.14 \times 10^{-1}$Å
= 0.414 Å = 0.0414 nm
$\begin{aligned} V =30 kV =30 \times 1000 & =30,000 V \\ & =3 \times 10^4 V\end{aligned}$
We know that : $E =h \nu= eV$
$\therefore$ From $\quad v_{\max }=\frac{ eV }{ h }$
Where $\quad h =6.63 \times 10^{-34} Js$
$e =1.6 \times 10^{-19} C$
$=7.24 \times 10^{18} Hz$
(b) Let minimum wavelength of X-rays $=\lambda_{\min }=$ ?
$\therefore \quad \lambda_{\min }=\frac{c}{v_{\max }}$
Substituting values
$\lambda_{\min }=\frac{3 \times 10^8}{7.24 \times 10^{18}}$
$=4.14 \times 10^{-11} m$
$=4.14 \times 10^{-11} \times 10^{10}$Å
$=4.14 \times 10^{-1}$Å
= 0.414 Å = 0.0414 nm
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