A charge particle of $$2 \mu C$$ accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is ______ $$\times 10^{-18}$$ kg.

Using kinetic energy from potential difference: $$K = qV$$

Using path radius in a magnetic field: $$R = \frac{\sqrt{2mK}}{qB} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$$

$$R^2 = \frac{2mV}{qB^2} \implies m = \frac{q B^2 R^2}{2V}$$

$$m = \frac{2 \times 10^{-6} \times (4 \times 10^{-3})^2 \times (3 \times 10^{-2})^2}{2 \times 100}$$

$$m = \frac{2 \times 10^{-6} \times (16 \times 10^{-6}) \times (9 \times 10^{-4})}{200}$$

$$m = \frac{288 \times 10^{-16}}{200} = 1.44 \times 10^{-16}\text{ kg} = 144 \times 10^{-18}\text{ kg}$$