Answer
$(7.07\times 10^{-8})^{\circ}$
Work Step by Step
The vertical distance between $n^{th}$ interference minimum and the center of the interference pattern is given by
$y_n=\frac{D}{d}(2n+1)\frac{\lambda}{2}$ ..................$(1)$
where $D$ is the distance between the slit and the viewing screen. $d$ is the slit separation.
For second minimum $n=1$, then eq. $(1)$ becomes
$y_2=\frac{3D\lambda}{2d}$
or, $\frac{y_2}{D}=\frac{3\lambda}{2d}$ ..................$(2)$
Let $\theta$ be the angle between the center of the pattern and the second minimum (to either side of the center).
$\therefore\;\;\tan\theta=\frac{y_2}{D}$ ..................$(3)$
Substituting eq. $(2)$ in eq. $(3)$, we get
$\tan\theta=\frac{3\lambda}{2d}$
or, $\theta=\tan^{-1}\Big(\frac{3\lambda}{2d}\Big)$ ..................$(4)$
Now the de Broglie wavelength of the proton is given by
$\lambda=\frac{h}{P}$ ..................$(5)$
Each with a speed of $v=0.9900c$. To find momentum $P$, we have to consider relativistic momentum, which is given by
$P=\gamma mv$, where $\gamma=\frac{1}{\sqrt {1-v^2/c^2}}$
Thus, eq. $(5)$ becomes,
$\lambda=\frac{h}{\gamma mv}$ ..................$(5)$
Substituting $(5)$ in eq. $(4)$, we get,
$\theta=\tan^{-1}\Big(\frac{3h}{2d\gamma mv}\Big)$ ..................$(6)$
For, $v=0.9900c$, $\gamma=\frac{1}{\sqrt {1-(0.99)^2}}=7.0888$
Substituting the given values in eq. $(6)$, we get
$\theta=\tan^{-1}\Big(\frac{3\times 6.63 \times 10^{-34}}{2\times 4\times 10^{-9}\times 7.0888\times 1.67\times 10^{-27}\times 0.99\times3\times 10^{8}}\Big)$
$\theta= (7.07\times 10^{-8})^{\circ}$
