Dual Nature of Matter

JEE Main level, numerical answer, medium

11 of 200

Question 11 of 200

A particle having electric charge 3×10−19C3 \times 10^{-19} \mathrm{C} and mass 6×10−27 kg6 \times 10^{-27} \mathrm{~kg} is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α×10−12 m\alpha \times 10^{-12} \mathrm{~m}. The value of α\alpha is ____\_\_\_\_ - (Take Planck's constant =6.6×10−34 J.s=6.6 \times 10^{-34} \mathrm{~J}. \mathrm{s})

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Answer: 10

The de Broglie wavelength is λ=hmv\lambda = \frac{h}{mv} where vv is found from energy conservation.
From qV=12mv2qV = \frac{1}{2}mv^2: v=2qVm=2×3×10−19×1.216×10−27=1.21×108=1.1×104v = \sqrt{\frac{2qV}{m}} = \sqrt{\frac{2 \times 3 \times 10^{-19} \times 1.21}{6 \times 10^{-27}}} = \sqrt{1.21 \times 10^8} = 1.1 \times 10^4 m/s.
Therefore: λ=6.6×10−346×10−27×1.1×104=6.6×10−346.6×10−23=10−11\lambda = \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 = 10×10−1210 \times 10^{-12} m.
Thus α=10\alpha = 10.

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