國際物理奧賽介紹-資料下載頁

【導讀】式從物體中放出的假設。在1905年，愛因斯坦發展。一顆不變的光量子。的光子聚在波前處形成的。本實驗的目的是測量。射的物體，同時它也是完全的輻射體。電磁波來說，黑體是完全吸收，沒有反射，真實物體不是“全黑”。率比就是該物體的發射率?的電磁波的功率密度為。值小時，即在圖F-1左方離極大值。確定c2，它正比于普朗克常數h。輻射物體為白熾燈A的。試管B裝有濾光液體，光敏電阻的電阻R與其所受光照E. 為LDR的特征之一，需由本。從上面可得出LDR的電阻R與鎢絲。金屬電阻率隨溫度的升高而增加。熱膨脹造成燈絲長度和截面的改變。以調節到的最小電壓至1V之間。斷開電位器與電池之間的導線。驗能夠達到的最小電壓。將所選擇的所有數據填在答卷紙的表中。LDR的材料在黑暗中是不導電的。這里b是常數,它僅與LDR的成分和幾何結構有關。理論上，理想的LDR的γ=1，然而，基于許。多因素的作用，在實際情況下γ<1。量R和E的關系來完成。在完成問題4的b)部分之前，不要

　　

【正文】 ts or quantities missing in any of the axes. pt Plot joined by sawtooth. pt 智浪教育 普惠英才文庫 c) p Appropriate data range pt Ranges from 0 to110 mV are acceptable. Ranges from 0 to 150 mV. pt Ranges including values beyond 150 mV. pt Fit made with less than 6 points. pt Computation of RB0 and ?RB0 . pt RB0 pt Least squares or equivalent numeric fit. Graphic determination only. pt Missing or incorrect units. pt Incorrect number of digits. pt If 10 ? RB0 13 ? pt ?RB0 pt Least squares or equivalent numeric fit. Graphic determination only. pt Error given without justification. pt RB0 and ?RB0 given with inconsistent number of digits. pt 智浪教育 普惠英才文庫 d) pt a value pt For any putational error in the use of RB0 or ?RB0 . pt ?a value pt Appropriate numeric method used. Error given without justification. pt a and ?a given with inconsistent number of digits. pt TASK 3. points ?0 pt For 585 nm ?0 595 nm . If ?0 595 nm or ?0 585 nm . pt ?0 given with precision better than 1 nm . pt ?? pt For 20 nm ? ??0 ? 30 nm . For ??0 20 nm or ??0 30 nm . pt ?0 and ??0 given with inconsistent number of digits. pt 智浪教育 普惠英才文庫 TASK 4. points a) pt 8 or more readings. From 6 to 8 readings. pt Less than 6 readings. pt Readings made with improper scales (200V, 20 A o R scale living too few digits). pt Units or quantities missing in each of the columns. pt Incorrect number of digits in each of V, I or R. pt 智浪教育 普惠英才文庫 b) 0 pt R and R’ pt R not the last reading made in the previous question.. pt R’ in the range between R and R R’ R or R’ R . pt ? and ?? pt ? 0 pt ? not puted . pt ? or ? . pt ? given with other than 2 digits. pt ?? 0 pt Computed using methods for indirect measurements or numeric methods. ?y given without justification. pt ? and ?? given with inconsistent number of digits. pt c) pt Missing or wrong equation. pt 智浪教育 普惠英才文庫 d) pt Column and values for 1/T missing. pt Column and values for ln R missing. pt Values of T or 1/T inconsistent with equation (6). pt lnR putation is incorrect. pt For each column with incorrect number of digits. pt Fit made with less than 12 points. pt 智浪教育 普惠英才文庫 e) pt Graphics pt Improper scale/aspect ratio obscuring data trend. (**) pt Units or quantities missing in any of the axes. pt Plot joined by saw tooth. pt More than 30 % of the data points are misaligned (*). pt h and ?h pt h pt Using least squares fit or equivalent numeric method. Graphically only. pt Missing or incorrect units . pt Incorrect number of digits. pt If h ?1034 Js or h ?1034 Js pt ?h pt Using least squares fit or equivalent numeric method. pt Graphically only. pt ?h given without justification. pt h and ?h given with inconsistent number of digits. pt 智浪教育 普惠英才文庫 Misaligned point: | 1 ( ) | 0 . 0 6iiy y x?? where ()iiy x m x b??** Improper scale/aspect ratio: ? ?? ? m inm inm a x ????yyl e n g t hxxl e n g t hm 醲NOTA BENE: For the case where the solution was obtained by inverse engineering, using the known value of the Planck constant: Each part of the question affected by this procedure will be given points. 智浪教育 普惠英才文庫 智浪教育 普惠英才文庫