Question

When the voltage applied to an x-ray tube increases from $V_1=10kV$ to $V_2=20kV$, the wavelength interval between $K_{\alpha }$ line and cut-off wavelength of continuous spectrum increases by a factor of $3$. Atomic number of the metallic target is

• A. $28$
• B. $30$
• C. $65$
• D. $66$

Answer 1

The correct option is B $30$

The cut-off wavelength when $V=V_1=10kV$ is ${\lambda }_1=\frac{hc}{e{V}_1}=\frac{(6.62\times {10}^{-34})\times (3\times {10}^8)}{(1.6\times {10}^{-19})\times (10\times {10}^3)}=12.41\times {10}^{-11}m$

The cut-off wavelength when $V=V_2=20kV$ is ${\lambda }_2=\frac{hc}{e{V}_2}=\frac{(6.62\times {10}^{-34})\times (3\times {10}^8)}{(1.6\times {10}^{-19})\times (20\times {10}^3)}=6.21\times {10}^{-11}m$

The wavelength corresponding $K_{\alpha }$ line is $\frac{1}{\lambda }=R(Z-1{)}^2(1/1^2-1/2^2)=\frac{3R(Z-1{)}^2}{4}$

According to question, $(\lambda -{\lambda }_2)=3(\lambda -{\lambda }_1)$

Dividing by $\lambda$ $\Rightarrow 1-\frac{{\lambda }_2}{\lambda }=3-3\frac{{\lambda }_1}{\lambda }$

or $\frac{1}{\lambda }(3{\lambda }_1-{\lambda }_2)=2$

or $\left(3/4\right)R(Z-1{)}^2(37.23-6.21){10}^{-11}=2$

Taking $R={10}^7{m}^{-1},(Z-1)=29.31$

or $Z=30$