The polymer prepared by a certain method exhibits a mole fraction and degree of polymerization that follow the following Poisson distribution.

$x_k=\frac{\lambda^k}{k!}e^{-\lambda}$

Where $x_k$ is the ratio of the amount of substance of polymer chains with a degree of polymerization of $k$ to the total amount of substance of polymer chains, $k$ is the degree of polymerization, and $\lambda$ is the reaction ratio of monomer to initiator. Assuming that the relative molecular mass of the polymer is $M_k=M_0+kM$. Then, for the following sets of parameters, the polymer molecular weight distribution that is the most uniform is:
以某方法制备得的聚合物表现出摩尔分数与聚合度会遵循如下的泊松分布。

$x_k=\frac{\lambda^k}{k!}e^{-\lambda}$

式中$x_k$ 为聚合度为$k$的聚合物链的物质的量与聚合物链总物质的量的比例，$k$ 为聚合度，$\lambda$为单体与引发剂的反应比例。假设该聚合物的相对分子质量为$M_k=M_0+kM$。则对于下面所给的几组参数中，聚合物分子量分布最均一的是：
All other options are incorrect
$\lambda=25,M_0=100,M=80$
$\lambda=25,M_0=100,M=100$
$\lambda=25,M_0=200,M=55$
$\lambda=25,M_0=200,M=75$
$\lambda=25,M_0=400,M=60$
$\lambda=25,M_0=400,M=76$
$\lambda=50,M_0=40,M=55$
$\lambda=50,M_0=40,M=65$
$\lambda=50,M_0=70,M=50$
$\lambda=50,M_0=70,M=60$
$\lambda=50,M_0=90,M=48$
$\lambda=50,M_0=90,M=57$
其他选项均不正确
$\lambda=25,M_0=100,M=80$
$\lambda=25,M_0=100,M=100$
$\lambda=25,M_0=200,M=55$
$\lambda=25,M_0=200,M=75$
$\lambda=25,M_0=400,M=60$
$\lambda=25,M_0=400,M=76$
$\lambda=50,M_0=40,M=55$
$\lambda=50,M_0=40,M=65$
$\lambda=50,M_0=70,M=50$
$\lambda=50,M_0=70,M=60$
$\lambda=50,M_0=90,M=48$
$\lambda=50,M_0=90,M=57$
The physical quantity that can reflect the molecular weight distribution of a polymer is the ratio of weight-average molecular weight to number-average molecular weight, $\frac{X_w}{X_n}$. Therefore, it is necessary to first derive the expression for $\frac{X_w}{X_n}$.

<Checkpoint>[$\frac{X_w}{X_n}$ can reflect the molecular weight distribution of the polymer](1)</Checkpoint>

According to the definition, we have $\frac{X_w}{X_n}=\frac{\sum n_k\sum n_kM_k^2}{(\sum n_kM_k)^2}=\frac{\sum x_kM_k^2}{(\sum x_kM_k)^2}$

<Checkpoint>[$\frac{X_w}{X_n}=\frac{\sum x_kM_k^2}{(\sum x_kM_k)^2}$](1)</Checkpoint>

Where $\sum x_kM_k=M_0\sum\frac{\lambda^k}{k!}e^{-\lambda}+M\sum k\frac{\lambda^k}{k!}e^{-\lambda}$

And we have $M_0\sum\frac{\lambda^k}{k!}e^{-\lambda}=M_0$

$M\sum k\frac{\lambda^k}{k!}e^{-\lambda}=M\lambda e^{-\lambda}\sum\frac{\lambda^{k-1}}{(k-1)!}=M\lambda$

Therefore, we have $\sum x_kM_k=M_0+M\lambda$

<Checkpoint>[$\sum x_kM_k=M_0+M\lambda$](1)</Checkpoint>

In addition, we have

$\sum x_kM_k^2=M_0^2\sum\frac{\lambda^k}{k!}e^{-\lambda}+2MM_0\sum k\frac{\lambda^k}{k!}e^{-\lambda}+M^2\sum k^2\frac{\lambda^k}{k!}e^{-\lambda}$

Where

$M_0^2\sum\frac{\lambda^k}{k!}e^{-\lambda}=M_0^2$

$2MM_0\sum k\frac{\lambda^k}{k!}e^{-\lambda}=2MM_0\lambda$

Specifically, we have

$\sum k^2\frac{\lambda^k}{k!}=\sum k\frac{\lambda^k}{(k-1)!}=\sum\frac{k}{k-1}\times\frac{\lambda^k}{(k-2)!}=\lambda^2\sum\frac{\lambda^k-2}{(k-2)!}+\lambda\sum\frac{\lambda^k-1}{(k-1)!}=e^{\lambda}(\lambda^2+\lambda)$

From this, we have

$M^2\sum k^2\frac{\lambda^k}{k!}e^{-\lambda}=M^2(\lambda^2+\lambda)$

<Checkpoint>[$M^2\sum k^2\frac{\lambda^k}{k!}e^{-\lambda}=M^2(\lambda^2+\lambda)$](0.5)</Checkpoint>

Therefore, we have

$\sum x_kM_k^2=M_0^2+2MM_0\lambda+M^2(\lambda^2+\lambda)$

<Checkpoint>[$\sum x_kM_k^2=M_0^2+2MM_0\lambda+M^2(\lambda^2+\lambda)$](1)</Checkpoint>

Therefore, we obtain

$\frac{X_w}{X_n}=1+\frac{\lambda M^2}{(M_0+M\lambda)^2}$

<Checkpoint>[$\frac{X_w}{X_n}=1+\frac{\lambda M^2}{(M_0+M\lambda)^2}$](2)</Checkpoint>

The value of $\frac{X_w}{X_n}$ can reflect the molecular weight distribution of the polymer; the smaller the value, the more uniform the distribution.

<Checkpoint>[The smaller the $\frac{X_w}{X_n}$, the more uniform the molecular weight distribution of the polymer](1)</Checkpoint>

Substituting the data in the question into the calculation, the values of $\frac{X_w}{X_n}-1$ corresponding to each option are $B=0.03628;C=0.03698;D=0.03049;E=0.03266;F=0.02493;G=0.02730;H=0.01943;I=0.01952;J=0.01893;K=0.01910;L=0.01858;M=0.01879$

The value of option L is the smallest, which is 0.01858
能够体现聚合物分子量分布的物理量是重均分子量与数均分子量之比$\frac{X_w}{X_n}$，因此需要先推导出$\frac{X_w}{X_n}$的表达式。

<Checkpoint>[$\frac{X_w}{X_n}$能够体现聚合物分子量分布](1)</Checkpoint>

根据定义有$\frac{X_w}{X_n}=\frac{\sum n_k\sum n_kM_k^2}{(\sum n_kM_k)^2}=\frac{\sum x_kM_k^2}{(\sum x_kM_k)^2}$

<Checkpoint>[$\frac{X_w}{X_n}=\frac{\sum x_kM_k^2}{(\sum x_kM_k)^2}$](1)</Checkpoint>

其中$\sum x_kM_k=M_0\sum\frac{\lambda^k}{k!}e^{-\lambda}+M\sum k\frac{\lambda^k}{k!}e^{-\lambda}$

并且有$M_0\sum\frac{\lambda^k}{k!}e^{-\lambda}=M_0$

$M\sum k\frac{\lambda^k}{k!}e^{-\lambda}=M\lambda e^{-\lambda}\sum\frac{\lambda^{k-1}}{(k-1)!}=M\lambda$

因此有$\sum x_kM_k=M_0+M\lambda$

<Checkpoint>[$\sum x_kM_k=M_0+M\lambda$](1)</Checkpoint>

此外还有

$\sum x_kM_k^2=M_0^2\sum\frac{\lambda^k}{k!}e^{-\lambda}+2MM_0\sum k\frac{\lambda^k}{k!}e^{-\lambda}+M^2\sum k^2\frac{\lambda^k}{k!}e^{-\lambda}$

其中

$M_0^2\sum\frac{\lambda^k}{k!}e^{-\lambda}=M_0^2$

$2MM_0\sum k\frac{\lambda^k}{k!}e^{-\lambda}=2MM_0\lambda$

特殊地，有

$\sum k^2\frac{\lambda^k}{k!}=\sum k\frac{\lambda^k}{(k-1)!}=\sum\frac{k}{k-1}\times\frac{\lambda^k}{(k-2)!}=\lambda^2\sum\frac{\lambda^k-2}{(k-2)!}+\lambda\sum\frac{\lambda^k-1}{(k-1)!}=e^{\lambda}(\lambda^2+\lambda)$

由此有

$M^2\sum k^2\frac{\lambda^k}{k!}e^{-\lambda}=M^2(\lambda^2+\lambda)$

<Checkpoint>[$M^2\sum k^2\frac{\lambda^k}{k!}e^{-\lambda}=M^2(\lambda^2+\lambda)$](0.5)</Checkpoint>

因此有

$\sum x_kM_k^2=M_0^2+2MM_0\lambda+M^2(\lambda^2+\lambda)$

<Checkpoint>[$\sum x_kM_k^2=M_0^2+2MM_0\lambda+M^2(\lambda^2+\lambda)$](1)</Checkpoint>

因此得到

$\frac{X_w}{X_n}=1+\frac{\lambda M^2}{(M_0+M\lambda)^2}$

<Checkpoint>[$\frac{X_w}{X_n}=1+\frac{\lambda M^2}{(M_0+M\lambda)^2}$](2)</Checkpoint>

$\frac{X_w}{X_n}$的值能够体现聚合物分子量分布，其值越小分布越均匀

<Checkpoint>[$\frac{X_w}{X_n}$越小聚合物分子量分布越均匀](1)</Checkpoint>

将题目中各个数据代入计算，得到各选项对应的$\frac{X_w}{X_n}-1$的值分别为$B=0.03628;C=0.03698;D=0.03049;E=0.03266;F=0.02493;G=0.02730;H=0.01943;I=0.01952;J=0.01893;K=0.01910;L=0.01858;M=0.01879$

选项L的值最小，为0.01858