See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Let  be the angular diffraction.  Sin  = n d  For small , sin  =  = n d  For 1st minima, n = 1   for first minima = d  Since, angular width of the central maximum is 3 2 2 4 10 d      rad   = d  2  103 m = 4  107 m = 4000 Å AITS-FT-V-PCM(Sol.)-JEE(Main)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 8 Chemistry PART – B SECTION – A