Chapter 24, Problem 64GP

To determine

The angles for the first-order diffraction line of sources.

Answer to Problem 64GP

The angle of hydrogen, ${\theta }_h=11.3°$

The angle of neon, ${\theta }_n=11.2°$

The angle of argon, ${\theta }_a=12.1°$

Explanation of Solution

Given:

The distance of grating lines, (d) $\text{d}\text{=}\text{3}\text{.00}{\text{×10}}^{\text{5}}\text{lines/}\text{m}$

Wavelength for three sources:

For hydrogen $\text{6}{\text{.56×10}}^{\text{-7}}\text{m}$ ,

For neon $\text{6}{\text{.50×10}}^{\text{-7}}\text{m}$ ,

For argon $\text{6}{\text{.97×10}}^{\text{-7}}\text{m}$ .

Order of diffraction, $\text{n}\text{=}\text{1}$

Formula used:

It is known that $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

Calculation:

The path difference is an integer multiple of the wavelength in constructive interference.

Rearrange the given formula: $\sin \theta =\frac{m\lambda }{d}$ . So, $d\sin \theta =m\lambda$

Calculation for Wavelength of hydrogen:

Substitute the given values in $d\sin {\theta }_h=m{\lambda }_h$

  $\frac{1}{3.0\times {10}^{5}lines/m}\sin {\theta }_h=6.56\times {10}^{-7}m$

  $\sin {\theta }_h=0.197$

  ${\theta }_h=11.3°$

Calculation for Wavelength of neon:

Substitute the given values in $d\sin {\theta }_n=m{\lambda }_n$

  $\frac{1}{3.0\times {10}^{5}lines/m}\sin {\theta }_n=6.50\times {10}^{-7}m$

  $\sin {\theta }_n=0.195$

  ${\theta }_n=11.2°$

Calculation for Wavelength of argon:

Substitute the given values in $d\sin {\theta }_a=m{\lambda }_a$

  $\frac{1}{3.0\times {10}^{5}lines/m}\sin {\theta }_a=6.97\times {10}^{-7}m$

  $\sin {\theta }_a=0.209$

  ${\theta }_a=12.1°$

Conclusion:

The angle of the first order of diffraction for hydrogen, neon and argon sources are $\text{11}\text{.3}°$ , $\text{11}\text{.2}°$ , $\text{12}\text{.1}°$ , respectively.