Question

If applied potential across the X-ray tube is made $\dfrac{2}{5}$ times then the minimum wavelength of X-rays is shifted by $1\mathop A\limits^0$ . Find the original values of applied potential and minimum wavelength

Answer 1

Hint : In order to solve this question, we are going to use the formula for the energy of an electron through a potential difference and also for the wave fo some wavelength, from these two equations and the information given, the wavelength is calculated and also the potential difference value.
The energy of an electron with charge $e$ and the applied potential difference $V$ as:
$E = eV$
Also the energy relation for the wave of frequency $\nu$ is
$E = h\nu$

Complete Step By Step Answer:
According to the Einstein’s theory, we have the formula for the energy of an electron with charge $e$ and the applied potential difference $V$ as:
$E = eV$
Also the energy relation for the wave of frequency $\nu$ is
$E = h\nu$
Equating the two, we get
$eV = h\nu$
Thus, the potential $V$ can be calculated as
$V = \dfrac{{h\nu }}{e}$
Where the values of the constants are
h = 6.626 \times {10^{ - 34}} \\\ e = 1.6 \times {10^{ - 19}} \\\ c = 3 \times {10^8} \\\
Putting the values of the constants, we get
V = \dfrac{{h\left( {\dfrac{c}{\lambda }} \right)}}{e} = \dfrac{{6.626 \times {{10}^{ - 34}}\left( {\dfrac{{3 \times {{10}^8}}}{\lambda }} \right)}}{{1.6 \times {{10}^{ - 19}}}} \\\ \Rightarrow V = \dfrac{{12367}}{\lambda } \\\
Now, as the potential difference is made $\dfrac{2}{5}$ times, then the wavelength shifts by unity
This implies
V \to \dfrac{2}{5}V \\\ \lambda \to \lambda + 1 \\\
Thus forming the equation for it, we get
$\dfrac{2}{5}V = \dfrac{{12367}}{{\lambda + 1}}$
Dividing the two equations for the potential differences, we get
\Rightarrow \dfrac{V}{{\dfrac{2}{5}V}} = \dfrac{{\dfrac{{12367}}{\lambda }}}{{\dfrac{{12367}}{{\lambda + 1}}}} \\\ \Rightarrow \dfrac{5}{2} = \dfrac{{\lambda + 1}}{\lambda } \\\
On solving this equation to get the value of the wavelength, we get
5\lambda = 2\lambda + 2 \\\ \Rightarrow \lambda = \dfrac{2}{3} = 0.667\mathop A\limits^0 \\\
Now, to find the potential difference, putting in the first equation, we get
$V = \dfrac{{12367}}{{0.667}} = 18665V$

Note :
It is to be noted that the step of taking the two different formulae for the energies of the particle is important that also shows the dual nature of the particle as two theories are supported here. After that utilizing the information for the variation of potential difference and wavelength is important.