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Two particles $\mathrm{A}$ and $\mathrm{B}$ of de - Broglie wavelengths $\lambda_1$ and $\lambda_2$ combine to form a particle $\mathrm{C}$. The process conserves momentum. Find the de-Broglie wavelength of the particle $\mathrm{C}$. (The motion is one - dimensional)
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Verified Answer
By de-Broglie wavelength
$$
\mathrm{P}=\frac{\mathrm{h} \lambda}{\lambda}, \quad \mathrm{P}_{\mathrm{A}}=\frac{\mathrm{h} \lambda}{\lambda_{\mathrm{A}}}, \quad \mathrm{P}_{\mathrm{B}}=\frac{\mathrm{h} \lambda}{\lambda_{\mathrm{B}}}, \mathrm{P}_{\mathrm{C}}=\frac{\mathrm{h} \lambda}{\lambda_{\mathrm{C}}}
$$
Given from conservation of momentum,
$$
\begin{aligned}
&\left|\mathrm{p}_{\mathrm{C}}\right|=\left|\mathrm{p}_{\mathrm{A}}\right|+\left|\mathrm{p}_{\mathrm{B}}\right| \\
&\Rightarrow \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{\mathrm{h}}{\lambda_{\mathrm{A}}}+\frac{\mathrm{h}}{\lambda_{\mathrm{B}}}\left[\because \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}} \Rightarrow \mathrm{p}=\frac{\mathrm{h}}{\lambda}\right] \\
&\Rightarrow \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{\mathrm{h} \lambda_{\mathrm{B}}+\mathrm{h} \lambda_{\mathrm{A}}}{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}} \\
&\Rightarrow \frac{\lambda_{\mathrm{C}}}{\mathrm{h}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\mathrm{h} \lambda_{\mathrm{A}}+\mathrm{h} \lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{C}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}+\lambda_{\mathrm{B}}}
\end{aligned}
$$
Case $\mathrm{I}$ if both $\mathrm{p}_{\mathrm{A}}$ and $\mathrm{p}_{\mathrm{B}}$ are positive.
then $\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$
Case II if both $\mathrm{p}_{\mathrm{A}}$ and $\mathrm{p}_{\mathrm{B}}$ are negative,
So, $\quad \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$
Case III if $\mathrm{p}_{\mathrm{A}}>0, \mathrm{p}_{\mathrm{B}} < 0$ i.e., $\mathrm{p}_{\mathrm{A}}$ is positive and $\mathrm{p}_{\mathrm{B}}$ is negative,
$$
\begin{aligned}
&\text { So, } \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{\mathrm{h}}{\lambda_{\mathrm{A}}}-\frac{\mathrm{h}}{\lambda_{\mathrm{B}}}=\frac{\left(\lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}\right) \mathrm{h}}{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}} \\
&\Rightarrow \lambda_{\mathrm{C}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}}
\end{aligned}
$$
Case IV, if $\mathrm{p}_{\mathrm{A}} < 0, \mathrm{p}_{\mathrm{B}}>0$, i.e., $\mathrm{p}_{\mathrm{A}}$ is negative and $\mathrm{p}_{\mathrm{B}}$ is positive,
$$
\begin{aligned}
&\text { So, } \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{-\mathrm{h}}{\lambda_{\mathrm{A}}}+\frac{\mathrm{h}}{\lambda_{\mathrm{B}}} \\
&\Rightarrow \quad=\frac{\left(\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}\right) \mathrm{h}}{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{C}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}}
\end{aligned}
$$
$$
\mathrm{P}=\frac{\mathrm{h} \lambda}{\lambda}, \quad \mathrm{P}_{\mathrm{A}}=\frac{\mathrm{h} \lambda}{\lambda_{\mathrm{A}}}, \quad \mathrm{P}_{\mathrm{B}}=\frac{\mathrm{h} \lambda}{\lambda_{\mathrm{B}}}, \mathrm{P}_{\mathrm{C}}=\frac{\mathrm{h} \lambda}{\lambda_{\mathrm{C}}}
$$
Given from conservation of momentum,
$$
\begin{aligned}
&\left|\mathrm{p}_{\mathrm{C}}\right|=\left|\mathrm{p}_{\mathrm{A}}\right|+\left|\mathrm{p}_{\mathrm{B}}\right| \\
&\Rightarrow \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{\mathrm{h}}{\lambda_{\mathrm{A}}}+\frac{\mathrm{h}}{\lambda_{\mathrm{B}}}\left[\because \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}} \Rightarrow \mathrm{p}=\frac{\mathrm{h}}{\lambda}\right] \\
&\Rightarrow \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{\mathrm{h} \lambda_{\mathrm{B}}+\mathrm{h} \lambda_{\mathrm{A}}}{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}} \\
&\Rightarrow \frac{\lambda_{\mathrm{C}}}{\mathrm{h}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\mathrm{h} \lambda_{\mathrm{A}}+\mathrm{h} \lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{C}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}+\lambda_{\mathrm{B}}}
\end{aligned}
$$
Case $\mathrm{I}$ if both $\mathrm{p}_{\mathrm{A}}$ and $\mathrm{p}_{\mathrm{B}}$ are positive.
then $\lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$
Case II if both $\mathrm{p}_{\mathrm{A}}$ and $\mathrm{p}_{\mathrm{B}}$ are negative,
So, $\quad \lambda_C=\frac{\lambda_A \lambda_B}{\lambda_A+\lambda_B}$
Case III if $\mathrm{p}_{\mathrm{A}}>0, \mathrm{p}_{\mathrm{B}} < 0$ i.e., $\mathrm{p}_{\mathrm{A}}$ is positive and $\mathrm{p}_{\mathrm{B}}$ is negative,
$$
\begin{aligned}
&\text { So, } \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{\mathrm{h}}{\lambda_{\mathrm{A}}}-\frac{\mathrm{h}}{\lambda_{\mathrm{B}}}=\frac{\left(\lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}\right) \mathrm{h}}{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}} \\
&\Rightarrow \lambda_{\mathrm{C}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}}
\end{aligned}
$$
Case IV, if $\mathrm{p}_{\mathrm{A}} < 0, \mathrm{p}_{\mathrm{B}}>0$, i.e., $\mathrm{p}_{\mathrm{A}}$ is negative and $\mathrm{p}_{\mathrm{B}}$ is positive,
$$
\begin{aligned}
&\text { So, } \frac{\mathrm{h}}{\lambda_{\mathrm{C}}}=\frac{-\mathrm{h}}{\lambda_{\mathrm{A}}}+\frac{\mathrm{h}}{\lambda_{\mathrm{B}}} \\
&\Rightarrow \quad=\frac{\left(\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}\right) \mathrm{h}}{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}} \Rightarrow \lambda_{\mathrm{C}}=\frac{\lambda_{\mathrm{A}} \lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}-\lambda_{\mathrm{B}}}
\end{aligned}
$$
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