The potential difference between the anode and the cathode of a vacuum tube is 2.0 x 10^2 V. Free electrons originate at the cathode with negligible speed and move to the anode. The speed with which the electrons arrive at the anode is?
(1/2) m v^2 = q V
m is electron mass
q is electron charge (assume positive)
To determine the speed with which the electrons arrive at the anode, we can use the equation for electric potential energy:
Potential energy (PE) = Charge (q) x Potential difference (V)
In this case, the potential difference is given as 2.0 x 10^2 V. Assuming a single electron, the charge (q) is the elementary charge, which is 1.6 x 10^-19 C.
We can rearrange the equation to solve for the speed of the electron:
PE = 1/2 x mass (m) x velocity^2
Since the electron starts with negligible speed at the cathode, the potential energy at the cathode is equal to the kinetic energy at the anode:
PE at the cathode = KE at the anode
Now, let's calculate the potential energy at the cathode:
PE at the cathode = q x V at the cathode
PE at the cathode = (1.6 x 10^-19 C) x (2.0 x 10^2 V)
Next, we equate it to the kinetic energy at the anode:
KE at the anode = 1/2 x m x v^2
Setting the potential energy at the cathode equal to the kinetic energy at the anode, we have:
(1.6 x 10^-19 C) x (2.0 x 10^2 V) = 1/2 x (mass of electron) x v^2
We know the mass of an electron is approximately 9.1 x 10^-31 kg. Substituting this value, we can solve for 'v'.