Question:
In a cathode ray experiment hydrogen gas was subjected to very low pressure and high voltage. The $\frac{e}{m}$ of cathode rays liberated was $-x~C~kg^{-1}$. In the second experiment helium gas was used under the same experimental conditions. What is the $\frac{e}{m}$ $(in~C~kg^{-1})$ of cathode rays liberated in the second experiment?}
In a cathode ray experiment hydrogen gas was subjected to very low pressure and high voltage. The $\frac{e}{m}$ of cathode rays liberated was $-x~C~kg^{-1}$. In the second experiment helium gas was used under the same experimental conditions. What is the $\frac{e}{m}$ $(in~C~kg^{-1})$ of cathode rays liberated in the second experiment?}
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Cathode rays are electrons. Their charge-to-mass ratio is always the same regardless of the gas inside the discharge tube.
Updated On: Jun 17, 2026
- $-2x$
- $-x$
- $\frac{x}{2}$
- $\frac{x^2}{2}$
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The Correct Option is B
Solution and Explanation
Concept:
Cathode rays are streams of electrons.
The value of
\[
\frac{e}{m}
\]
for electrons is independent of the gas present in the discharge tube.
Step 1: Identify the nature of cathode rays.
Cathode rays always consist of electrons. Their properties do not depend on the gas used.
Step 2: Apply Thomson's conclusion.
Whether hydrogen or helium is used, \[ \frac{e}{m} \] remains unchanged. Hence, \[ \left(\frac{e}{m}\right)_{\text{He}} = -x \]
Step 3: State the answer.
\[ \boxed{-x} \]
Step 1: Identify the nature of cathode rays.
Cathode rays always consist of electrons. Their properties do not depend on the gas used.
Step 2: Apply Thomson's conclusion.
Whether hydrogen or helium is used, \[ \frac{e}{m} \] remains unchanged. Hence, \[ \left(\frac{e}{m}\right)_{\text{He}} = -x \]
Step 3: State the answer.
\[ \boxed{-x} \]
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