A particle having electric charge $$3\times 10^{-19}$$ C and mass $$6 \times 10^{-27}$$ kg is accelerated by applying an electric Potential of 1.21 V. Wavelength of the matter wave
associated with the particle is $$\alpha \times 10^{-12}m$$. The value of $$\alpha$$ is __________. (Take Planck's constant = $$6.6 \times 10^{-34}$$ J.s. )

Charge $$q = 3 \times 10^{-19}$$ C, mass $$m = 6 \times 10^{-27}$$ kg, potential $$V = 1.21$$ V.

KE gained: $$KE = qV = 3 \times 10^{-19} \times 1.21 = 3.63 \times 10^{-19}$$ J

$$\frac{1}{2}mv^2 = 3.63 \times 10^{-19}$$

$$v^2 = \frac{2 \times 3.63 \times 10^{-19}}{6 \times 10^{-27}} = 1.21 \times 10^{8}$$

$$v = 1.1 \times 10^4$$ m/s

de Broglie wavelength: $$\lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{6 \times 10^{-27} \times 1.1 \times 10^4} = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}} = 10^{-11}$$ m

Wait, $$\lambda = 10^{-11} = 10 \times 10^{-12}$$ m. So $$\alpha = 10$$.

The answer is 10.