Find the maximum frequency produced by \(30\;\text{kV}\) electrons.
Answer
In this question we are asked to find ‘maximum frequency’ and ‘minimum wavelength’ produced by \(30\;\text{kV}\) electrons. This is nothing, but the frequency and wavelength produced by the given energy of electrons. Energy is directly proportional to frequency, so the frequency obtained from the given energy \((\)which is the maximum possible energy\()\) is the maximum frequency. Similarly, Energy is inversely proportional to wavelength, so the wavelength obtained from the given energy is the minimum wavelength.
Given that, Electron potential,
\(V=30\mathit{\ kV}=30\times 10^4V\)
We know that, Energy,
\(E=\mathit{qV}\)
Where, \(V\) is the potential and \(q\) is the charge.
Here, \(q=e\)
Hence, Electron energy,
\(E=30\times 10^4\mathit{eV}\)
We Know that,
\(E=\mathit{h\nu }\)
Where, \(\nu \) is the frequency and \(h=6.626\times 10^{-34}\mathit{Js}\)
Therefore, the maximum frequency of electron,
\(\nu =\frac E h=\frac{30\times 10^4}{6.626\times 10^{-34}}\)
\(=7.24\times 10^{18}\mathit{Hz}\)
Hence, \(7.24\times 10^{18}\mathit{Hz}\) is the maximum frequency of \(X\) -rays.
Given that, Electron potential,
\(V=30\mathit{\ kV}=30\times 10^4V\)
We know that, Energy,
\(E=\mathit{qV}\)
Where, \(V\) is the potential and \(q\) is the charge.
Here, \(q=e\)
Hence, Electron energy,
\(E=30\times 10^4\mathit{eV}\)
We Know that,
\(E=\mathit{h\nu }\)
Where, \(\nu \) is the frequency and \(h=6.626\times 10^{-34}\mathit{Js}\)
Therefore, the maximum frequency of electron,
\(\nu =\frac E h=\frac{30\times 10^4}{6.626\times 10^{-34}}\)
\(=7.24\times 10^{18}\mathit{Hz}\)
Hence, \(7.24\times 10^{18}\mathit{Hz}\) is the maximum frequency of \(X\) -rays.
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