If the velocity of electron increases, its specific charge

# If the velocity of electron increases, its specific charge

1. A

increases

2. B

decreases

3. C

remains constant

4. D

may increase or decrease

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### Solution:

$\text{Mass\hspace{0.17em}of\hspace{0.17em}electron}=\frac{\text{rest\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}electron}}{\sqrt{\text{1}-\frac{{\text{V}}^{\text{2}}}{{\text{C}}^{\text{2}}}}}$
V = Velocity of electron
C = Velocity of light
As velocity of electron increases,$\frac{{\text{V}}^{\text{2}}}{{\text{C}}^{\text{2}}}$   increases and $\left(\text{1}-\frac{{\text{V}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)$  decreases as denominator decreases mass of electron increases (rest mass electron is a constant) and thus  e/m (specific charge) decreases

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