A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
1 answer:
a) Zinc (work function: 4.3 eV)
The equation for the photoelectric effect is:
(1)
where
is the energy of the incident photon, with
h = Planck constant
c = speed of light
= wavelength
= work function of the metal
K = maximum kinetic energy of the photoelectrons emitted
The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,
So we can rewrite (1) as
where we have:
V = 1.93 V
e is the electron charge
First of all, let's find the energy of the incident photon:
Converting into electronvolts,
And now we can solve eq.(1) to find the work function of the metal:
so, the metal is most likely zinc, which has a work function of 4.3 eV.
b) The stopping potential is still 1.93 V
Explanation:
The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.
Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.
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The correct answer is
Rahul’s weight
In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.
Using formulas, Rahul's weight is equal to
where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).
Explanation:
Given that,
Number of turns in the coil, N = 52
Original are of the coil,
It is stretched to have no area in 0.100 s, t = 0.1 s
The magnetic field strength, B = 1.2 T
To find,
The magnitude and the direction of average induced emf.
Solution,
The Faraday law gives the magnitude and the direction of induced EMF in the coil. Its formula is given by :
E = 171.6 volts
The direction of induced emf is given by Lenz law. It opposes the direction due to which it is produced. It is in counterclockwise direction.
Answer:
U₁ > U₂
Explanation:
By definition, the capacitance of a capacitor, can be expressed as follows:
,
where Q is the charge on one of the terminals of the capacitor, and V, the potential difference between them.
If the capacitors are of the parallel plate type, applying Gauss ' law to a gaussian surface defined as a pill box parallel to the surface of one of the plates (half outside the plate, half inside it) , the capacitance can be expressed as follows:
where ε is the dielectric constant of the material that fills the space between plates, A is the area of one of the plates (assuming a constant surface charge density) and d is the distance between the plates.
If both capacitors have the same charge Q, as they are identical, the potential difference V must be the same for both capacitors also.
So, the electric potential energy, for both capacitors, can be written as follows:
Now, if after being charged and disconnected (which means that the charge must keep the same), C₂ is filled with a dielectric, looking to the equation (2), we can see that the value of the capacitance, will increase, proportionally to the value of the dielectric constant of the dielectric.
If we analyze the expression (3) for the electric potential energy stored in the capacitor, as Q₂ = Q₁, and being C₂ = ε*C₁, this means that the electric potential energy U₂ must be lower than U₁, in the same factor ε, as follows:
⇒ U₁ > U₂