Given that charge of the electron is 1.6 x10^-19 C,What is the energy gained by the cathode ray particle when a voltage of 800 volts is applied between the electrodes of a cathode ray tube
Well, putting my math hat on, we know that the formula to calculate the energy is given by E = qV, where E is the energy, q is the charge, and V is the voltage.
So, plugging in the values, we get E = (1.6 x 10^-19 C) x (800 V).
Now, that's quite a small charge times a large voltage, which means the energy gained will be a tiny amount.
Let me crunch the numbers for you. *grabs calculator*
Okay, after some calculations, the energy gained by the cathode ray particle is... *drumroll* ... I think I need to sneeze... Ah-choo! Sorry, moving on.
The energy gained by the cathode ray particle is approximately 1.28 x 10^-16 joules. It may be a small amount, but hey, every little bit counts, right?
To find the energy gained by the cathode ray particle, we can use the equation:
Energy = Charge × Voltage
Given:
Charge of electron (q) = 1.6 × 10^-19 C
Voltage (V) = 800 V
Substituting the values into the equation:
Energy = (1.6 × 10^-19 C) × (800 V)
Calculating the expression:
Energy = 1.28 × 10^-16 J
Therefore, the energy gained by the cathode ray particle when a voltage of 800 volts is applied is approximately 1.28 × 10^-16 Joules.
To calculate the energy gained by the cathode ray particle, you can use the formula:
Energy = Charge x Voltage
Given that the charge of the electron is 1.6 x 10^-19 C and the voltage applied is 800 volts, you can substitute these values into the formula:
Energy = (1.6 x 10^-19 C) x (800 volts)
To evaluate this expression, you can multiply the numbers together:
Energy = 1.28 x 10^-16 J
Therefore, the energy gained by the cathode ray particle when a voltage of 800 volts is applied is 1.28 x 10^-16 Joules.