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An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10 \mathrm{keV}$ and the second with $100 \mathrm{keV}$. Which is faster, the electron or proton? Obtain the ratio of their speeds.
Take mass of electron $=9.11 \times 10^{-31} \mathrm{~kg}$. Mass of proton $=1.67 \times 10^{-27} \mathrm{~kg}$ and $1 \mathrm{keV}=1.6 \times 10^{-16} \mathrm{~J}$.
Take mass of electron $=9.11 \times 10^{-31} \mathrm{~kg}$. Mass of proton $=1.67 \times 10^{-27} \mathrm{~kg}$ and $1 \mathrm{keV}=1.6 \times 10^{-16} \mathrm{~J}$.
Solution:
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Verified Answer
Kinetic energy, $E_e=\frac{1}{2} m_e v_e^2=10 \times 1.6 \times 10^{-16}$;
$$
E_p=\frac{1}{2} m_p v_p^2=100 \times 1.6 \times 10^{-16} .
$$
Dividing, $\frac{m_e v_e^2}{m_p v_p^2}=\frac{1}{10}$
$$
\begin{aligned}
\Rightarrow \frac{v_e}{v_p} &=\sqrt{\frac{1}{10} \cdot \frac{m_p}{m_e}}=\sqrt{\frac{1.67 \times 10^{-27}}{10 \times 9.11 \times 10^{-31}}} \\
&=10^2 \sqrt{\frac{1.67}{91.1}}=13.53
\end{aligned}
$$
$\therefore \quad v_e=13.53 v_{\mathrm{p}} \Rightarrow$ the electron is faster than the proton.
$$
E_p=\frac{1}{2} m_p v_p^2=100 \times 1.6 \times 10^{-16} .
$$
Dividing, $\frac{m_e v_e^2}{m_p v_p^2}=\frac{1}{10}$
$$
\begin{aligned}
\Rightarrow \frac{v_e}{v_p} &=\sqrt{\frac{1}{10} \cdot \frac{m_p}{m_e}}=\sqrt{\frac{1.67 \times 10^{-27}}{10 \times 9.11 \times 10^{-31}}} \\
&=10^2 \sqrt{\frac{1.67}{91.1}}=13.53
\end{aligned}
$$
$\therefore \quad v_e=13.53 v_{\mathrm{p}} \Rightarrow$ the electron is faster than the proton.
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