Answer:
B. Maximum velocity of ejected electrons.
Explanation:
The ejection of electrons form a metal surface when the metal surface is exposed to a monochromatic electromagnetic wave of sufficiently short wavelength or higher frequency (or equivalently, above a threshold frequency), which leads to the enough energy of the wave to incident and get absorbed to the exposed surface emits electrons. This phenomenon is known as the photoelectric effect or photo-emission.
The minimum amount of energy required by a metal surface to eject an electron from its surface is called work function of metal surface.
The electrons thus emitted are called photo-electrons.
The current produced as a result is called photo electricity.
Energy of photon is given by:
where:
h = Planck's constant
frequency of the incident radiation.
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
Answer:
1.1ohms
Explanation:
According to ohms law E = IR
If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5
I = 0.36A (This will be the load current).
Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.
Voltage drop = 2.2V - 1.8V = 0.4V
Then we calculate the internal resistance using ohms law.
According to the law, V = Ir
V= voltage drop
I is the load current
r = internal resistance
0.4 = 0.36r
r = 0.4/0.36
r = 1.1 ohms
The capacitance is defined as the maximum charge stored in a capacitor, Q, divided by the voltage applied, V:
The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula:
The energy stored in a capacitor is
E = (1/2) · (capacitance) · (voltage)²
E = (1/2) · (6 x 10⁻⁶ F) · (12 V)²
E = (3 x 10⁻⁶ F) · (144 V²)
<em>E = 4.32 x 10⁻⁴ Joule</em>
(That's 0.000432 of a Joule)
