An x-ray tube is operating at 30 kV then the minimum wavelength of the x-rays coming out of the tube is:

1. 1.24 $\stackrel{0}{A}$

2. 0.413 $\stackrel{0}{A}$ 

3. 0.124 $\stackrel{0}{A}$ 

4. 0.13 $\stackrel{0}{A}$

Bragg's law for X-rays is:

1. dsin$\theta$ = 2n$\lambda$ 

2. 2dsin$\theta$ = n$\lambda$

3. nsin$\theta$ = 2$\lambda$d

4. None of these

The minimum wavelength of X-rays produced by electrons accelerated by a potential difference of V volt is equal to 

1. $\frac{eV}{hc}$ 

2. $\frac{eh}{cV}$ 

3. $\frac{hc}{eV}$ 

4. $\frac{cV}{eh}$

The graph between the square root of the frequency of a specific line of the characteristic spectrum of X-rays and the atomic number of the target will be:

1.
2.
3.
4.

The wavelength of $K_{\alpha }$ X-rays for lead isotopes $P{b}^{208},<mspace\; width="1em"></mspace>P{b}^{206},<mspace\; width="1em"></mspace>P{b}^{204}<mspace\; width="1em"></mspace>are<mspace\; width="1em"></mspace>{\lambda }_1,<mspace\; width="1em"></mspace>{\lambda }_2<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>{\lambda }_3$ respectively. Then

1. ${\lambda }_1={\lambda }_2>{\lambda }_3$   

2. ${\lambda }_1>{\lambda }_2>{\lambda }_3$

3. ${\lambda }_1<{\lambda }_2<{\lambda }_3$ 

4. ${\lambda }_2={\lambda }_1={\lambda }_3$