In the photoelectric effect, the energy given by the incoming photon is used partially to extract the electron from the metal (work function) and the rest is converted into kinetic energy of the electron:
where
hf is the energy of the photon, with h being the Planck constant and f the frequency of the photon
is the work function
K is the kinetic energy of the electron
When K=0, we have the minimum energy required to extract the electron from the metal, so the equation becomes
(1)
If we convert the work function of gold into Joules:
We can re-arrange eq.(1) to find the minimum energy of the photon:
Answer:
a. 4.733 × 10⁻¹⁹ J = 2.954 eV b i. yes ii. 0.054 eV = 8.651 × 10⁻²¹ J
Explanation:
a. Find the energy of the incident photon.
The energy of the incident photon E = hc/λ where h = Planck's constant = 6.626 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/s and λ = wavelength of light = 420 nm = 420 × 10⁻⁹ m
Substituting the values of the variables into the equation, we have
E = hc/λ
= 6.626 × 10⁻³⁴ Js × 3 × 10⁸ m/s ÷ 420 × 10⁻⁹ m
= 19.878 × 10⁻²⁶ Jm ÷ 420 × 10⁻⁹ m
= 0.04733 × 10⁻¹⁷ J
= 4.733 × 10⁻¹⁹ J
Since 1 eV = 1.602 × 10⁻¹⁹ J,
4.733 × 10⁻¹⁹ J = 4.733 × 10⁻¹⁹ J × 1 eV/1.602 × 10⁻¹⁹ J = 2.954 eV
b. i. Is this energy enough for an electron to leave the atom
Since E = 2.954 eV is greater than the work function Ф = 2.9 eV, an electron would leave the atom. So, the answer is yes.
ii. What is its maximum energy?
The maximum energy E' = E - Ф = 2.954 - 2.9
= 0.054 eV
= 0.054 × 1 eV
= 0.054 × 1.602 × 10⁻¹⁹ J
= 0.08651 × 10⁻¹⁹ J
= 8.651 × 10⁻²¹ J
Answer:
Moving the magnet away from the center of the loop with its south pole facing the center of the loop.
Explanation:
Electromagnetic induction is due to a rapidly changing magnetic field, or loop area. The poles of the magnet induce current in the loop but in the opposite direction, depending on the direction of their relative motion. An approaching north pole will induce an anticlockwise current in the loop, while an approaching south pole will do the reverse. To get the galvanometer to flicker in the same direction as of that when the north pole was approaching, we move the magnet away from the center of the loop with its south pole facing the center of the loop.
The magnitude of the force is 18.6 N
Explanation:
The work done by a force on an object is given by
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the block in this problem, we have
is the work done
d = 1.35 m is the displacement of the block
is the angle between the force and the displacement
Solving for F, we find the magnitude of the force:
Learn more about work here:
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To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:
Where,
= Mass of each object
= Initial Velocity of Each object
= Final Velocity
Our values are given as
Replacing we have that
Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s
