Question

When the voltage applied to an x-ray tube is increased from $10kV$ to $20kV$, the wavelength interval between the ${{K}_{\alpha }}$ line and the short wave cut off of the continuous X-ray spectrum increases by a factor $3$. Find the atomic number of the element of which the tube anti cathode is made. (Rydberg’s constant, $R={{10}^{7}}{{m}^{-1}}$)

Answer 1

Hint: We have to use the relationship of wavelength in terms of planck's constant, speed of light, energy and applied voltage. Relation of wavelength with atomic number will also be used in order to solve the problem. Atomic number of an element is defined as the number of protons present in the atom.

Formula Used:
We are using the following two formulae to solve the given problem:-
$\dfrac{1}{\lambda }=R{{(Z-1)}^{2}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$ and $\lambda =\dfrac{hc}{eV}$

Complete answer:
From the above given question we have the following parameters with us:-
Initial voltage, ${{V}_{1}}=10kV$
Final voltage, ${{V}_{2}}=20kV$
Rydberg’s constant, $R={{10}^{7}}{{m}^{-1}}$
We have to find the atomic number of the element, $Z$.
We will first, find the wavelength, ${{\lambda }_{1}}$ for the initial voltage by using the following relation:-
${{\lambda }_{1}}=\dfrac{hc}{e{{V}_{1}}}$………………… $(i)$
Where $h=6.626\times {{10}^{-34}}Js$, and called Planck's constant and$c$is the speed of light. Charge of electron is given by $e$ and value of $e=1.6\times {{10}^{-19}}C$. Putting parameters in $(i)$ we get
${{\lambda }_{1}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 10}$
$\Rightarrow $${{\lambda }_{1}}=1.24\overset{o}{\mathop{A}}\,$ ……………….. $(ii)$
Now doing same process for the final voltage we get:-
${{\lambda }_{2}}=\dfrac{hc}{e{{V}_{2}}}$…………….. $(iii)$
Putting the values of parameters in equation $(iii)$we get
${{\lambda }_{2}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}\times 20}$
${{\lambda }_{2}}=6.21\overset{o}{\mathop{A}}\,$……………. $(iv)$
Now, wavelength $\lambda $ in terms of shell number ${{n}_{1}}$ and ${{n}_{2}}$, corresponding to ${{K}_{\alpha }}$ (${{K}_{\alpha }}$ line results when an electron makes transition to the innermost K-shell ) is given as follows:-
$\dfrac{1}{\lambda }=R{{(Z-1)}^{2}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
$\dfrac{1}{\lambda }=R{{(Z-1)}^{2}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right]$……………. $(v)$ (we put ${{n}_{1}}=1$ and ${{n}_{2}}=2$)
Solving further we get
$\dfrac{1}{\lambda }=\dfrac{3R}{4}{{(Z-1)}^{2}}$…………………….. $(vi)$
Now the increasing factor is $3$. It means that
$(\lambda -{{\lambda }_{2}})=3(\lambda -{{\lambda }_{1}})$……………. $(vii)$
Putting the values of $(ii)$ and $(iv)$ in $(vii)$, we get
$\left( \lambda -6.21 \right)=3\left( \lambda -1.24 \right)$
$\Rightarrow -2\lambda =2.49$
$\Rightarrow \lambda =1.245\overset{o}{\mathop{A}}\,$……………. $(viii)$ (Neglecting minus sign)
Putting these parameters in $(vi)$we get
$\dfrac{1}{1.245}={{\left( Z-1 \right)}^{2}}\times {{10}^{7}}\times \dfrac{3}{4}$
Solving further we get
${{\left( Z-1 \right)}^{2}}=1.245\times \dfrac{3}{4}\times {{10}^{7}}$
$\Rightarrow (Z-1)=\sqrt{0.934\times {{10}^{7}}}$
Solving further we get
$Z=31$, The given atom is Gallium.

Note:
We should use the values of constants properly and without any confusion. Concept of increasing factor should be used to relate the wavelengths at different voltages. Whenever, we get the value of the atomic number in decimal form then we should round-off it to get the proper whole number.