NTSE Stage 2 SAT 2016 Question Paper

Question : 28 of 100

Marks: +1, -0

In a neon gas discharge tube, every second

4.8 ×10^18

Ne+ ions move towards the right through a croos-section of the tube, while ‘n’ electrons move to the left in the same time. If the current in the tube is 1.12 amperes towards the right, n is equal to (given e =

1.6 ×10^{–19}

coulomb)

$1.8×10^{–18}$
$2.2 ×10^{–18}$
$2.4 ×10^{–19}$
$2.8 ×10^{–19}$

Validate

Solution:

(4.8×10^18+\n)1.6×10^{-19}=1.12

(4.8×10^18+\n)=1.12/{1.6×10^{-19}}

4.8×10^{18}+\n=7×10^{18}

\n=7×10^{18}-4.8×10^{18}=2.2×10^{18}

So, n is equal to

2.2×10^{18}

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