A wire carries a current of 4.1 A. How many electrons per second are passing any cross sectional area of the wire? Enter your an
1 answer:
To solve this problem it is necessary to apply the concepts related to Current and Load.
The current in terms of the charge of an electron can be expressed as
Where,
q = Charge
t = time
At the same time the Charge is the amount of electrons multiplied by the amount of these, that is
q = ne
Replacing in the first equation we have to
Clearing n,
Here the time is one second then
Therefore the number of electrons per second are passing any cross sectional area of the wire are
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Answer:
λ = 482.05 nm
Explanation:
The diffraction phenomenon and the diffraction grating is described by the expression
d sin θ = m λ
where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction
in this case they indicate the distance between slits, the angle and the order of diffraction
λ = d sin θ / m
let's calculate
λ = 1.00 10⁻⁶ sin 74.6 / 2
λ = 4.82048 10⁻⁷ m
Let's reduce to nm
λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)
λ = 482.05 nm
Answer:
Explanation:
As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron
So here we have to find the work done by electric field on moving electron
So we have
now the distance moved by the electron is given as
so we have
now we have to convert it into keV units
so we have