Answer
a. $53.8^{\circ}$.
b. 0.19 nm.
Work Step by Step
a. Use equation 25–10, which applies to x-ray diffraction.
$$m\lambda=2dsin\phi$$
Apply it to each situation. The spacing d is the same, as is the wavelength.
$$\frac{m_1}{m_2}=\frac{sin \phi_1}{sin \phi_2}$$
$$\phi_2=sin^{-1}(\frac{m_2}{m_1} sin \phi_1)= sin^{-1}(\frac{2}{1} sin 23.8^{\circ})=53.8^{\circ}$$
b. Find the wavelength using the m = 1, first-order data.
$$m\lambda=2dsin\phi$$
$$\lambda=\frac{2dsin\phi}{m}=\frac{2(0.24nm) sin 23.8^{\circ}}{1}=0.19nm$$
