解開蔗糖水解的秘密
本研究利用偏振片、量角器為刻度盤、雷射光為光源，及照度計為偵測器，組裝一個簡易且可靠的旋光度計。我們利用單位時間旋光度的變化量當作反應速率，來測量蔗糖的水解速率，同時求出蔗糖水解反應的反應級數、速率常數(k)。利用糖類的旋光度具有加成性之特性，找出不同混合比例時的旋光度，追蹤實際蔗糖水解的每個狀態，找出最後平衡狀態，同時將蔗糖水解平衡結果顯示，旋光度與濃度有線性關係，而蔗糖水解反應對蔗糖而言為一級反應。接著，我們在蔗糖水溶液中加入不同種類的酸，探討催化劑的種類與蔗糖水解反應速率的關係。 In this research, in order to measure the optical rotation accurately without expensive equipments or complex process, we assembled a polarimeter by ourselves. With simple materials which can be found in ordinary senior high school laboratories, including a calibrated scale, a simple Luxmeter, a laser as the photo source, and other side devices. The Polarimeter ended up operating fluently and accurately. We put the laser under a tube, which has two pieces of polar screens on the top of it and on the bottom of it, ,and put a luxmeter just above the tube. When we slowly rotate the polar screen on the top, the figure shown on the luxmeter changes. By numerical analysis, we can get information about the hydrolysis of polarized substance. Secondary, we measured the optical rotation of glucose, fructose, malt sugar, galactose, and sucrose to get their specific rotation. Then we measured the optical rotation of sucrose every five minutes. By doing this, we could keep track of the hydrolysis rate of sucrose, figure out the order of reaction, and the rate constant (k) and the equilibrium constant (K). Thirdly, we used different kinds of acids into sucrose solution as the catalyst, and observed the effect. The result showed that hydrochloric acid is a better catalyst to this reaction than sulfuric acid and nitric acid. The polarimeter of this research can be used in science education of junior and senior high school. By teaching students to assemble and operate the self-made polarimeter, students can know better about optical rotation and polarized substance. Also, the interest in this experiement will add to students’ motivation to do science research.
移動棋子問題的致勝策略
We consider a game played with chips on a strip of squares. The squares are labeled, left to right, with 1, 2, 3, . . ., and there are k chips initially placed on distinct squares. Two players take turns to move one of these chips to the next empty square to its left. In this project, we study four different games according to the following \r rules: Game A: the player who places a chip on square 1 wins;Game B: the player who places a chip on square 1 loses;Game C: the player who finishes up with chips on 12 . . . k wins;Game D: the player who finishes up with chips on 12 . . . k loses. After studying the cases k = 3, 4,5 and 6 for Game A and the relation among these four games, we are led to discover the winning strategy of each game for any positive integer k. The strategies of Games A, B and C are closely related through a forward or backward shifting in position. We also found that such strategies are similar to the type of Nim game that awards the player taking the last chip. Game D is totally different from the rest. To solve this game, we investigate the Nim game that declares the player taking the last chips loser. Amazingly, the strategies of two Nim games can be concisely linked by two equations. Through these two Nim games, we not only find the winning strategy of Game D but also the precise relation between Game D and all others.\r 去年我研究一個遊戲：有一列n個的方格中，從左至右依序編號為1,2,3,....n。在X1個、第X2個、第X3個格子中各放置一個棋子。甲乙二個人按照下列規則輪流移動棋子：\r 一、甲乙兩個人每次只能動一個棋子(三個棋子中任選一個)。遊戲開始由甲先移動動棋子。二、甲乙兩個人每次移動某一個棋子時，只能將這個棋子移至左邊最近的空格(若前面連續有P個棋時可以跳過前面的P個棋子而且只能跳一次)，而且每個方格中最多只能放一個棋子。\r 研究這個遊戲問題時，我討論四種不同"輸贏結果"的規定：甲乙兩個人中，A誰先將三個棋子中任意一個棋子移到第一個方格，誰就是贏家。B誰先將三個棋子中任意一個棋子移到第一個方格，誰就是輸家。C誰先不能再移動任何棋子，誰就是輸家。D誰先不能再移動任何棋子，誰就是贏家。\r 當"輸贏結果"的規定採用ABCD時─我們稱為遊戲ABCD。今年我將把這個遊戲問題中棋子的個數由三個推廣到一般K個情形之後，再繼續研究遊戲的致勝策略，同時也將研究遊戲ABCD之間的關係。