A charged particle is suspended in equilibrium in a uniform verticle field electric field of intensity 20000 V-M- If mass of the particle is 9-6 - 10-16 kg- the charge on it and excess number of electrons on the particle are respectively g - 10 m-s2

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De Broglie's Hypothesis

A charged par...

Question

A charged particle is suspended in equilibrium in a uniform verticle field electric field of intensity 20000 V/M. If mass of the particle is $9.6 \times 10^{- 16}$ kg,, the charge on it and excess number of electrons on the particle are respectively ( $g = 10 m / s^{2}$ )

4.8 \times 10^{- 19} C

3

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5.8 \times 10^{- 19} C

4

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3.8 \times 10^{- 19} C

2

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2.8 \times 10^{- 19} C

1

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Solution

The correct option is A $4.8 \times 10^{- 19} C$ , $3$
Electric field of intensity $= 20, 000 V / m$
The mass of particle $= 9.6 \times 10^{- 16} k g$
$g = 10 m / s^{2}$
$∴$ Charge $= \frac{20, 000}{9.6 \times 10^{- 16}} = 4.8 \times 10^{- 19} C$
Now, the charge on it and excess number of electrons on the particle are $= 4.8 \times 10^{- 19} - 1.6 \times 10^{- 19}$
$\approx 5 - 2$
$= 3$

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A charged particle is suspended in equilibrium in a uniform verticle field electric field of intensity 20000 V/M. If mass of the particle is 9.6×10−16 kg,, the charge on it and excess number of electrons on the particle are respectively (g=10m/s2)

The correct option is A 4.8×10−19C, 3Electric field of intensity =20,000V/mThe mass of particle =9.6×10−16kgg=10m/s2∴ Charge =20,0009.6×10−16=4.8×10−19CNow, the charge on it and excess number of electrons on the particle are =4.8×10−19−1.6×10−19 ≈5−2 =3

A charged particle is suspended in equilibrium in a uniform verticle field electric field of intensity 20000 V/M. If mass of the particle is $9.6 \times 10^{- 16}$ kg,, the charge on it and excess number of electrons on the particle are respectively ( $g = 10 m / s^{2}$ )