In the experiment, the relative intensities of three different mass signals of dichlorodibromomethane in the mass spectrum were measured as $M(240): 38.67 \%, M+2: 100 \%, M+4: 88.64 \%$.  

It is known that the main isotopes of chlorine are $^{35} \mathrm{Cl}$ and $^{37} \mathrm{Cl}$ with an abundance ratio of approximately $3: 1$, and the main isotopes of bromine are $^{79} \mathrm{Br}$ and $^{81} \mathrm{Br}$ with an abundance ratio of approximately $1: 1$.  

Which of the following statements is correct (a calculation error of less than 0.5% is considered acceptable):
实验中测得二氯二溴化碳在质谱中三
种不同质量数的信号相对强度为 $M(240): 38.67 \%, M+2: 100 \%, M+4: 88.64 \%$, 

已知氯元素的核素主要为丰度比约为 $3: 1$ 的 $^{35} \mathrm{Cl}$ 和 $^{37} \mathrm{Cl}$, 溴元素的核素主要为丰度比约为 $1: 1$的 $^{79} \mathrm{Br}$ 和 $^{81} \mathrm{Br}$。

下列说法正确的是（计算误差小于0.5%即算正确）：
All other options are incorrect
The abundance of $^{35} \mathrm{Cl}$ is $74.2\%$
The abundance of $^{37} \mathrm{Cl}$ is $25.2\%$
The abundance of $^{79} \mathrm{Br}$ is $51.6\%$.
The abundance of $^{81} \mathrm{Br}$ is $49.7\%$.
The relative signal intensity of $M+6$ is $34.5\%$.
The relative signal intensity of $M+8$ is $1.57\%$.
Let the abundance of $\mathrm{^{13}C}$ be $1\%$, then the relative signal intensity of $M+1$ theoretically detected in this experiment is approximately $0.4\%$.
其他选项均不正确
$^{35} \mathrm{Cl}$的丰度为$74.2\%$
$^{37} \mathrm{Cl}$的丰度为$25.2\%$
$^{79} \mathrm{Br}$的丰度为$51.6\%$
$^{81} \mathrm{Br}$的丰度为$49.7\%$
$M+6$的相对信号强度为$34.5\%$
$M+8$的相对信号强度为$1.57\%$
令$\mathrm{^{13}C}$的丰度计为$1\%$，则该实验理论上检测到$M+1$的相对信号强度约为$0.4\%$
Let $^{35} \mathrm{Cl}$ be event $A$ with probability $x$; $^{79} \mathrm{Br}$ be event $B$ with probability $y$.

$$
P(M)=P(A A B B)=x x y y
$$

$$
P(M+2)= P(\bar{A} A B B+A \bar{A} B B+A A \bar{B} B+A A B \bar{B})=2 x(1-x) y y+2 x x y(1-y) 
$$

$$
P(M+4)= P(\bar{A} \bar{A} B B+\bar{A} A \bar{B} B+\bar{A} A B \bar{B}+A \bar{A} \bar{B} B+A \bar{A} B \bar{B}+A A \bar{B}\bar{B}) 
$$
$$
= (1-x)^{2} y y+4 x(1-x) y(1-y)+x x(1-y)^{2}
$$

According to the problem,

$$
P(M): P(M+2): P(M+4)=0.3867: 1: 0.8864
$$

Finally, we can solve to obtain $x=75.8 \%, y=50.7 \%$.

Therefore, the abundance of ${ }^{35} \mathrm{Cl}$ is $75.8 \%$, and the abundance of ${ }^{37} \mathrm{Cl}$ is $24.2 \%$;

<Checkpoint>[The abundance of ${ }^{35} \mathrm{Cl}$ is $75.8 \%$, and the abundance of ${ }^{37} \mathrm{Cl}$ is $24.2 \%$;](2)</Checkpoint>

The abundance of ${ }^{79} \mathrm{Br}$ is $50.7 \%$, and the abundance of ${ }^{81} \mathrm{Br}$ is $49.3 \%$;

<Checkpoint>[The abundance of ${ }^{79} \mathrm{Br}$ is $50.7 \%$, and the abundance of ${ }^{81} \mathrm{Br}$ is $49.3 \%$;](2)</Checkpoint>

$$
P(M+6)=P(A \bar{A}\bar{B}\bar{B}+\bar{A}\bar{A} B \bar{B}+\bar{A} A \bar{B}\bar{B}+\bar{A}\bar{A} \bar{B}B)
$$

$$
=2 x(1-x)(1-y)^{2}+2(1-x)^{2} y(1-y)=30.23 \%
$$

<Checkpoint>[$P(M+6)=30.23 \%$](1)</Checkpoint>
 
$$
P(M+8)=P(\overline{A A B B})=(1-x)^{2}(1-y)^{2}=1.43 \%
$$

<Checkpoint>[$P(M+8)=1.43 \%$](1)</Checkpoint>

Therefore, options B-G are all incorrect.

The ratio of the relative signal intensity of $M+1$ to that of $M$ is equal to the ratio of the abundances of $^{13}C$ to $^{12}C$:

$\frac{P(M+1)}{P(M)}=\frac{1\%}{1-1\%}=\frac{x}{38.67\%}$

Solving gives $x≈0.4\%$

<Checkpoint>[$P(M+1)=0.4\%$](1)</Checkpoint>

Option H is correct.
将 $^{35} \mathrm{Cl}$ 设为事件 $A$, 概率为 $x$; $^{79} \mathrm{Br}$ 设为事件 $B$, 概率为 $y$。

$$
P(M)=P(A A B B)=x x y y
$$

$$
P(M+2)= P(\bar{A} A B B+A \bar{A} B B+A A \bar{B} B+A A B \bar{B})=2 x(1-x) y y+2 x x y(1-y) 
$$

$$
P(M+4)= P(\bar{A} \bar{A} B B+\bar{A} A \bar{B} B+\bar{A} A B \bar{B}+A \bar{A} \bar{B} B+A \bar{A} B \bar{B}+A A \bar{B}\bar{B}) 
$$
$$
= (1-x)^{2} y y+4 x(1-x) y(1-y)+x x(1-y)^{2}
$$

根据题意, 

$$
P(M): P(M+2): P(M+4)=0.3867: 1: 0.8864
$$

最终可以解出, $x=75.8 \%, y=50.7 \%$ 。

所以 ${ }^{35} \mathrm{Cl}$ 的丰度为 $75.8 \%,{ }^{37} \mathrm{Cl}$ 的丰度为 $24.2 \%$ ;

<Checkpoint>[${ }^{35} \mathrm{Cl}$ 的丰度为 $75.8 \%,{ }^{37} \mathrm{Cl}$ 的丰度为 $24.2 \%$ ;](2)</Checkpoint>

${ }^{79} \mathrm{Br}$ 的丰度为 $50.7 \%,{ }^{81} \mathrm{Br}$ 的丰度为 $49.3 \%$;

<Checkpoint>[${ }^{79} \mathrm{Br}$ 的丰度为 $50.7 \%,{ }^{81} \mathrm{Br}$ 的丰度为 $49.3 \%$;](2)</Checkpoint>

$$
P(M+6)=P(A \bar{A}\bar{B}\bar{B}+\bar{A}\bar{A} B \bar{B}+\bar{A} A \bar{B}\bar{B}+\bar{A}\bar{A} \bar{B}B)
$$

$$
=2 x(1-x)(1-y)^{2}+2(1-x)^{2} y(1-y)=30.23 \%
$$

<Checkpoint>[$P(M+6)=30.23 \%$](1)</Checkpoint>
 
$$
P(M+8)=P(\overline{A A B B})=(1-x)^{2}(1-y)^{2}=1.43 \%
$$

<Checkpoint>[$P(M+8)=1.43 \%$](1)</Checkpoint>

因此，选项B-G均错误。

$M+1$的相对信号强度与$M$的相对信号强度之比即为$^{13}C$与$^{12}C$的丰度之比：

$\frac{P(M+1)}{P(M)}=\frac{1\%}{1-1\%}=\frac{x}{38.67\%}$

求得$x≈0.4\%$

<Checkpoint>[$P(M+1)=0.4\%$](1)</Checkpoint>

选项H正确。