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The voltage applied between the cathode and anode of an X-ray tube is 18kV. Calculate the minimum wavelength of the X-rays produced.

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Last updated date: 17th Apr 2024
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Answer
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Hint: When the voltage is applied between cathode and anode a ray composed of electrons is produced in an X-ray tube. To find the minimum wavelength of the ray, use the concept of the kinetic energy of the electron. we can compare the maximum energy of the electro with that of a photon and can find the minimum wavelength produced in the X-ray tube.

Complete step by step solution:
Step 1: The voltage applied is given in kilovolt. First, convert it into volt. Therefore, $V = 18kV = 18 \times {10^3}V$ .
We know that the maximum kinetic energy is given by $K{E_{\max }} = eV$ , where $e$ is the value of the charge of an electron.
Step 2: But the energy required to give speed to an electron is given by $\dfrac{{hc}}{{{\lambda _{\min }}}}$ therefore
$\therefore eV = \dfrac{{hc}}{{{\lambda _{\min }}}}$
$ \Rightarrow {\lambda _{\min }} = \dfrac{{hc}}{{eV}}$, where $h$ is a planck constant. Its value is $6.63 \times {10^{ - 34}}Js$. $e$ is the charge of an electron $1.6 \times {10^{ - 19}}coulombs$. Here $c$ is the speed of light in a vacuum. Its value is $3 \times {10^8}m/s$.
Step 3: put the value in the above formula
$\therefore {\lambda _{\min }} = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}} \times 18 \times {{10}^3}}}$
Now simplify the above equation and we get
$\therefore {\lambda _{\min }} = 6.88 \times {10^{ - 11}}m$ .
Step 4: The wavelength of the X-rays is very small generally. Therefore we can convert the above result into angstrom by multiplying with ${10^{10}}$ . Therefore,
$\therefore {\lambda _{\min }} = 6.88 \times {10^{ - 11}} \times {10^{10}}A°$
$ \Rightarrow {\lambda _{\min }} = 0.688A°$

Hence the minimum wavelength of the X-rays will be $0.688A°$.

Note: While solving any numerical problem we should keep all the given values in the same unit system.
Since the energy cannot be destroyed therefore the energy required to make an electron leave the metal is equal to the energy given by a photon. The kinetic energy will be equal to the energy of a photon.