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# 有關 transmittance(穿透率)的問題<急>

A student measures the visible absorption spectrum for a 1.50 x 10-4 M solution of potassium dichromate in a 0.0025 M sulfuric acid solution. The absorbance of this solution at a wavelength of 366 nm was found to be 0.130. Calculate the percent transmittance (to 3 sig figs), %T, at this wavelength.

### 2 Answers

- 光弟Lv 71 decade agoFavorite Answer
比爾定律(Beer law): A = εbc = O.DA: 吸收值(absorbance) = O.D(光學密度) = E(消光值) = log(Po/P) 無單位Po/P = T, Transmittance, 穿透率, 傳輸值ε: molar absorption coefficient, 莫耳吸收係數, 單位: dm3 /(mol*cm) = L/mol cmb: cell thickness = 1 cmc: 容積莫耳濃度, 單位: mol / L = mol / dm3本題中 A = 0.130 = log Po/P T(transmittance) = Po/P = 0.886Percent transmittance = 88.6%

- 1 decade ago
應該要定義一下Po與P,ㄧ般而言

Po是穿透前的功率 P是穿透後的功率

所以 Po/P = T 應該改成 P/Po = T

O.D=log(Po/P) =log(1/T) 或是多ㄧ個負號

O.D=-1og(P/Po)=-log(T)

還有

本題中 A = 0.130 = log Po/P T(transmittance) = Po/P = 0.886

Percent transmittance = 88.6%

Po/P= 0.886 帶回 log Po/P=-0.052566278 不等於 O.D=0.130 ?