An electron has a mass of 9.1×10−31‌kg. It revolves round the nucleus in a circular orbit of radius 0.529×10−10m at a speed of 2.2×106ms−1. The magnitude of its angular momentum is
To find the magnitude of the angular momentum of the electron, we can use the formula for the angular momentum of a particle moving in a circular orbit: L=mvr where: m is the mass of the electron, 9.1×10−31‌kgv is the speed of the electron, 2.2×106ms−1 r is the radius of the orbit, 0.529×10−10m Now, let's substitute these values into the formula: L=9.1×10−31‌kg×2.2×106ms−1×0.529×10−10m First, we will calculate the product: 2.2×106×0.529×10−10 2.2×0.529=1.1638 Combining the exponents of 10 : 106×10−10=10−4 So the product is: 1.1638×10−4 Now calculate:
9.1×10−31‌kg×1.1638×10−4 Calculating the product of the coefficients: 9.1×1.1638 9.1×1.1638=10.57758 Combining this with the exponent of 10 : 10.57758×10−31×10−4=10.57758×10−35 We express this final value in scientific notation: 1.057758×10−34Kgm2s−1 Rounding this to three significant figures: 1.06×10−34Kgm2s−1 Therefore, the correct answer is: Option A: 1.06×10−34Kgm2s−1