Question

An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 33.5 cm away.

(a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field?

Answer 1

6.3445×10⁻¹⁶ m

Step-by-step explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

`E=(mv^2)/(2)\\\Rightarrow v=\sqrt{(2E)/(m)}\\\Rightarrow v=\sqrt{(2* 2470* 1.6* 10^(-19))/(9.11* 10^(-31))}\\\Rightarrow v=29455356.08671\ m/s`

Deflection by Earth's Gravity

`\Delta =\frac {gt^2}{2}`

Now, Time = Distance/Velocity

`\Delta =\frac {g(s^2)/(v^2)}{2}\\\Rightarrow \Delta =(9.81(0.335^2)/(29455356.08671^2))/(2)\\\Rightarrow \Delta =6.3445* 10^(-16)\ m`

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m