An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir
2 answers:
A photoelectric cell is an electronic device which is used to convert light energy into electric energy.The operation of this device is based on photoelectric effect.
Light of suitable frequency i.e greater or equal to threshold frequency will fall on the cathode maintained at negative potential.The electron emission will take place and these electrons are drifted towards the anode which is at positive potential.
Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.
We know that the radiation having long wavelength has least energy as energy and wavelength are inversely proportional to each other.
Here h is the Planck's constant,c is the velocity of light.
Here we have been given red light and blue light.
In the visible spectrum of radiation, the red light has longer wavelength than all other colors of light.Hence blue light has more energy as it's wavelength is less as compared to red light.
Hence, the blue light will activate the most and red the least.
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Answer:
Force is repulsive hence direction of force is away from wire
Explanation:
The first thing will be to draw a figure showing the condition,
Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.
Answer:
at y=6.29 cm the charge of the two distribution will be equal.
Explanation:
Given:
linear charge density on the x-axis,
linear charge density of the other charge distribution,
Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.
Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.
<u>we know, the electric field due to linear charge is given as:</u>
where:
linear charge density
r = radial distance from the center of wire
permittivity of free space
Therefore,
∴at y=6.29 cm the charge of the two distribution will be equal.
The electrical potential energy of a charge q located at a point at potential V is given by
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector: