Given that,
the distance between the two slits "d=2.7\\times 10^{-6}m"
first order maxima "\\theta=9.5"
a) We know that "\\sin\\theta=1.22\\dfrac{\\lambda}{d}"
"\\Rightarrow \\lambda=\\dfrac{d\\sin\\theta}{1.22}=\\dfrac{2.7\\times10^{-6}\\sin 9.5^\\circ m}{1.22}"
"\\Rightarrow \\lambda=\\dfrac{2.7\\times0.16\\times10^{-6}}{1.22}=0.35\\times 10^{-6}m"
b)
"\\dfrac{mv^2}{2}=\\dfrac{hc}{\\lambda}-\\phi"
"\\Rightarrow 0.51eV=\\dfrac{hc}{\\lambda}-2.28eV"
"\\Rightarrow \\dfrac{hc}{\\lambda}=0.51eV+2.28eV=2.79eV"
"\\Rightarrow \\lambda=\\dfrac{6.6\\times 10^{-34}\\times3\\times10^{3}}{2.79\\times1.6\\times10^{-19}}"
"\\Rightarrow \\lambda=4.43\\times10^{-12}m"
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