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Two particles A and B have wave length \(\lambda_A\) = 5 x 10-10m and \(\lambda_B\)= 10 x 10-10 m. Find their frequency, wave number and energies. Which has more penetrating power and why? |
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Answer» For particle A \(\lambda_A\) = 5 x 10-10 m Frequency vA = \(\frac{c}{\lambda_A}\) = \(\frac{3\times10^8\,m/s}{5\times10^{-10}m}\) = 0.6 x 1018 = 6.0 × 1017 sec−1 Wave number \(\bar{v_A}\) = \(\frac{1}{\lambda_A}\) = \(\frac{1}{5\times10^{-10}m}\) = 0.2 × 1010 m−1 = 2.0 × 109 m−1 = 2.0 × 107 cm−1 Energy EA = \(\frac{hc}{\lambda_A}\) = hvA = 6.626 × 10−34js × 6.0 × 1017 s−1 = 39.756 × 10−17 J = 3.9756 × 10−16J For Particle B \(\lambda_B\) = 10 x 10-10 m Frequency vB = \(\frac{c}{\lambda_B}\) = \(\frac{3\times10^8m/sec}{10\times10^{-10}m}\) = 0.3 x 1018 = 3.0 × 1017 sec−1 Wave number \(\bar{v_B}\) = \(\frac{1}{\lambda_B}\) = \(\frac{1}{10\times10^{-10}m}\) = 0.1 × 10−10 m−1 = 1.0 × 109 m−1 = 1.0 × 107 cm−1 Energy EB = \(\frac{hc}{\lambda_B}\) = hvB = 6.626 × 10−34Js x 3.0 × 1017 s−1 = 19.878 × 10−17 J = 1.9878 × 10−16J Since the energy of particle A is more than energy of particle B, so particle A has more penetrating power. |
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