The cathode rays generated in a vacuum tube are accelerated through a potential difference of \(300 \space \text V\).

Calculate the maximum velocity of the electron.

[Mass of electron, \(m = 9.1 \times 10^{-31} \space \text {kg}\), electron charge, \(e = 1.6 \times 10^{-19} \space \text C\)]

(A)  \(1.00 \times 10^4 \space \text m \text s^{-1}\)

(B)  \(1.01 \times 10^5 \space \text m \text s^{-1}\)

(C)  \(1.02 \times 10^6 \space \text m \text s^{-1}\)

(D)  \(1.03 \times 10^7 \space \text m \text s^{-1}\)

The above statement refers to \(\underline {\hspace {3.5cm}}\).

In a vacuum tube, the resulting cathode rays are accelerated through a potential difference of \(2.5 \space \text {kV}\).

Calculate the electrical potential energy of the cathode ray initially.

[Electron charge, \(e = 1.6 \times 10^{-19} \space \text C\)]

(A)  \(2.0 \times 10^{-16} \space \text J\)

(B)  \(2.0 \times 10^{-18} \space \text J\)

(C)  \(4.0 \times 10^{-16} \space \text J\)

(D)  \(2.0 \times 10^{-18} \space \text J\)

Which particle is cathode ray?

Source: SPM 2013

The diagram shows a Maltese Cross Tube.

The shadow of the green Maltese Cross is shaped in a position similar to the shaded by filament light.

What conclusion can be drawn from the above experiment?