Question

Consider the following particles:
(A) Proton,
(B) α-particle,
(C) Electron.
For charge to mass ratio of the above particles, we can say:

• (C) $>$ (A) $>$ (B)
• (A) $<$ (C) $<$ (B)
• (B) = (A) $>$ (C)
• (C) = (B) $>$ (A)

Answer 1

The Correct Option is A

To determine the charge to mass ratio of the given particles, we analyze each particle individually:

• Proton (A): Has a charge of +e (where e is the elementary charge) and a mass of approximately 1 atomic mass unit (amu).
• α-particle (B): Consists of 2 protons and 2 neutrons, thus has a total charge of +2e and a mass of approximately 4 amu.
• Electron (C): Has a charge of -e and a mass of approximately 1/1836 amu (significant due to small mass resulting in a high charge to mass ratio).

The charge to mass ratio is given as:

• Proton: \( \frac{+e}{1\, \text{amu}} \)
• α-particle: \( \frac{+2e}{4\, \text{amu}} = \frac{+e}{2\, \text{amu}} \)
• Electron: \( \frac{-e}{1/1836\, \text{amu}} = -1836\frac{e}{\text{amu}} \) (high magnitude, ignore sign for comparison)

Therefore, the order of charge to mass ratio by magnitude is: (C) > (A) > (B)

Answer 2

1. Proton (A)
2. α-particle (B)
3. Electron (C)

Charge-to-Mass Ratio Analysis:

Particle	Charge (q)	Mass (m)	q/m Ratio
Proton (A)	\(+e\)	\(m_p\)	\(\frac{e}{m_p}\)
α-particle (B)	\(+2e\)	\(4m_p\)	\(\frac{2e}{4m_p} = \frac{e}{2m_p}\)
Electron (C)	\(-e\)	\(m_e \approx \frac{m_p}{1836}\)	\(\frac{e}{m_e} \approx 1836\frac{e}{m_p}\)

Comparison Results:

From the calculations:

\[ \left(\frac{q}{m}\right)_C \approx 1836\left(\frac{q}{m}\right)_A \]

\[ \left(\frac{q}{m}\right)_A = 2\left(\frac{q}{m}\right)_B \]

Thus the order of charge-to-mass ratios is:

\[ \text{Electron (C)} \gg \text{Proton (A)} > \alpha\text{-particle (B)} \]

Correct Option:

\(\boxed{(1)\ (C) > (A) > (B)}\)