Đáp án:

Câu 1:

a,

$$I=I_o.10^{-\varepsilon.C.l}$$ 

$$\Rightarrow T=\dfrac{I}{I_o}=10^{-\varepsilon.C.l}$$ 

Khi $$T=0,5, C=1,28.10^{-4}M, l=10mm=1cm$$:

$$\varepsilon=\dfrac{-\log 0,5}{1,28.10^{-4}.1}=2351,8 M^{-1}.cm^{-1}$$ 

Khi $$C=1M, l=1cm$$, hệ số hấp thụ phân tử là:

$$\varepsilon_{\mu}=A=2351,8.1.1=2351,8$$ 

b,

$$T=10^{-\varepsilon.C.l}=10^{-2351,8.1,28.10^{-4}.2.1}=0,25=25\%$$

$$A=-\log T=0,602$$ 

Câu 2:

$$A=\epsilon.C.I$$ 

$$A_1=A_2\Rightarrow \epsilon.C_1.l_1=\epsilon.C_2.l_2$$ 

Vậy $$C_1l_1=C_2l_2$$ hay $$C_1=\dfrac{C_2l_2}{l_1}$$