n ammonia molecule is NH, and is shown in the figure The hydrogen atoms (mu = 1u, e atomic mass unit u 1.66 x 10- kg) are at the corners of an equilateral triangle that forms the base of a pyramid The lengths of the sides of the base are a- 1.6 A, and the nitrogen atom (mN- 14 u) is at the apex, A0.37 A vertically above the centre of the plane of the triangle. Find the height of the centre of mass of the ammonia molecule above the plane formed by its hydrogen atoms, ie above the base of the pyramid. Note you do not have to convert from u to kg (A) 0.13 A (8) 0.22 A (C) 0.30 A (D) 0 43 A (E) 0.49 A F) 0.51 A G) 0.62 A H) 0.71 A

n ammonia molecule is NH, and is shown in the figure The hydrogen atoms (mu = 1u, e atomic mass unit u 1.66 x 10- kg) are at the corners of an equilateral triangle that forms the base of a pyramid The lengths of the sides of the base are a- 1.6 A, and the nitrogen atom (mN- 14 u) is at the apex, A0.37 A vertically above the centre of the plane of the triangle. Find the height of the centre of mass of the ammonia molecule above the plane formed by its hydrogen atoms, ie above the base of the pyramid. Note you do not have to convert from u to kg (A) 0.13 A (8) 0.22 A (C) 0.30 A (D) 0 43 A (E) 0.49 A F) 0.51 A G) 0.62 A H) 0.71 A