At 150 degree Celsius, a mixture of 40 wt % Sn and 60 wt % Pb present, forming phases of alpha and beta. Chemical composition of Sn at each phase : CO (overall) : 40 %, CA (alpha) : 11 %, CB (beta) : 99 %. (a) State 2 reasons for the existences of alpha and beta phases for the mixture of Sn – Pb at 150 degree Celsius. (b) By using Lever Rule, calculate the weight fraction of each phase for alpha, WA = Q / (P Q) and beta, WB = P / (P Q) where Q = CB – CO and P = CO – CA.
Answer / kang chuen tat (malaysia - pen
(a) Reasons : (1) atomic radii difference; (2) different crystal structure. (b) P Q = CO – CA CB – CO = CB – CA, then WA = Q / (P Q) = (CB – CO) / (CB – CA) = (99 – 40) / (99 – 11) = 0.6705. WB = P / (P Q) = (CO – CA) / (CB – CA) = (40 – 11) / (99 – 11) = 0.3295. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS 61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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Question 99 - (a) The quantum number m is given by m = -s, -s + 1. If s = 0.5, find the values of m. (b) | T > = (cos T) | V > + (sin T) | H >. The V and H states form a basis for all polarizations. Let cos T = 0.8. (i) If (sin T)(sin T) + (cos T)(cos T) = 1, find the value of sin T. (ii) For | T > = a | V > + b | H >, where a x a represents the probability of | V > and b x b represents the probability of | H >. Which one is more abundant, | V > or | H >? (iii) Find the value of T without using any mathematical tools.
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