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Do all the electrons that absorb a photon come out as photoelectrons?
No, not all electrons that absorb a photon are ejected as photoelectrons. Only electrons that gain energy equal to or greater than the material's work function can escape. Others lose energy through collisions or remain bound within the material. For more visit here: https://www.tiwariacademy.com/ncRead more
No, not all electrons that absorb a photon are ejected as photoelectrons. Only electrons that gain energy equal to or greater than the material’s work function can escape. Others lose energy through collisions or remain bound within the material.
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A body of mass 1 kg is thrown upwards with a velocity of 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? [Take g = 10 m/s²]
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height. 1. Initial kinetic energy (K.E.) when the body is thrown: K.E. = (1/2) * m * v² K.E. = (1/2) * 1 kg * (20 m/s)² K.E. = (1/2) * 1 * 400 K.E. = 200 J. 2. PotentialRead more
To find the energy lost due to air friction, we first calculate the initial kinetic energy and the potential energy at the maximum height.
1. Initial kinetic energy (K.E.) when the body is thrown:
K.E. = (1/2) * m * v²
K.E. = (1/2) * 1 kg * (20 m/s)²
K.E. = (1/2) * 1 * 400
K.E. = 200 J.
2. Potential energy (P.E.) at the maximum height (h = 18 m):
P.E. = m * g * h
P.E. = 1 kg * 10 m/s² * 18 m
P.E. = 180 J.
3. The energy lost to air friction is given by the difference between the initial kinetic energy and the potential energy at the maximum height:
See lessEnergy lost = K.E at the start – P.E.
Energy lost = 200 J – 180 J
Energy lost = 20 J.
The coefficient of restitution e for a perfectly elastic collision is
The coefficient of restitution, denoted by 'e', describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e, eRead more
The coefficient of restitution, denoted by ‘e’, describes how elastic the collision between two bodies is. When the two bodies collide and go back into their original shape without any changes in momentum or kinetic energy after a perfectly elastic collision, it follows that, by definition of e,
e=relative velocity after collision/ relative velocity before collision
For the perfectly elastic collision, e equals 1.
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Two identical balls A and B collide head on elastically. If velocities of A and B, before the collision are +0.5 m/s and -0.3 m/s respectively, then their velocities, after the collision, are respectively
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved. 1. Let the initial velocities be: Velocity of A before collision (uA) = +0.5 m/s Velocity of B before collision (uB) = -0.3 m/Read more
To find the velocities of two identical balls A and B after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is also conserved.
1. Let the initial velocities be:
Velocity of A before collision (uA) = +0.5 m/s
Velocity of B before collision (uB) = -0.3 m/s
2. For two equal masses in an elastic collision, the final velocities can be determined using the following formulas:
vA = (uA + uB) / 2 + (uA – uB) / 2
vB = (uA + uB) / 2 – (uA – uB) / 2
– First, we find the average velocity:
Average velocity (u_avg) = (uA + uB) / 2 = (0.5 – 0.3) / 2 = 0.1 / 2 = 0.05 m/s.
Now, we compute the change in velocities:
Change in velocity = uA – uB = 0.5 – (-0.3) = 0.5 + 0.3 = 0.8 m/s.
3. Using the equations above:
– Final velocity of A (vA):
vA = 0.05 + 0.4 = 0.45 m/s to be corrected depending on the direction.
– Final velocity of B (vB):
vB = 0.05 – 0.4 = -0.35 m/s to be corrected depending on the direction.
4. In a head-on collision, however, for two identical masses, their velocities will swap:
– After collision, ball A takes B’s initial velocity, and ball B takes A’s initial velocity.
– Thus, the final velocities will be:
– vA = -0.3 m/s (B’s initial velocity)
– vB = +0.5 m/s (A’s initial velocity)
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Does the ‘stopping potential’ in photoelectric emission depend upon (i) the intensity of the incident radiation in a photocell? (ii) the frequency of the incident radiation?
(i) The stopping potential does not depend on the intensity of the incident radiation, as intensity affects the number of photoelectrons, not their energy. (ii) The stopping potential depends on the frequency of the incident radiation because higher frequency photons impart more energy to the emitteRead more
(i) The stopping potential does not depend on the intensity of the incident radiation, as intensity affects the number of photoelectrons, not their energy.
(ii) The stopping potential depends on the frequency of the incident radiation because higher frequency photons impart more energy to the emitted electrons, requiring greater potential to stop them.
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