## An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.7 × 10-16 N acts on it. The

Question

An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.7 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 22 mm horizontally.

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1 year 2021-08-31T01:49:54+00:00 1 Answers 4 views 0

1. Explanation:

GIven data :

Speed V = 1.1×10⁷ m/s

Force F = 3.7×10⁻¹⁶N

Mass of electron = m = 9.11×10⁻³¹Kg

Horizontal Distance X = 2.2×10⁻²m

Vertical distance Y = ?

Solution:

As we know that,

F =ma

where a is the acceleration of electron.

(3.7×10⁻¹⁶) = (9.11×10⁻³¹) a

a = 4.06×10¹⁴m/s² upward.

∵opposing weight of electron is negligible.

now we find how long it takes the electron to move 22mm horizontally.

X =V×T

T = X/V

T =2.2×10⁻²/1.1×10⁷

= 2×10⁻⁹s

now we can find out the vertical distance Y.

Y = 0.5aT²

= (0.5)(4.06×10¹⁴)(2×10⁻⁹)²

= 8.12×10⁻⁴m