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A photosensitive metallic surface has a work function, hv0. If photons of energy 2 hv0 fall on this surface, the electrons come out with a maximum velocity of 4 × 106 m/s. When the photon energy is increased to 5 hv0, then maximum velocity of photoelectrons will be
- a)2 × 107 m/s
- b)2 × 106 m/s
- c)8 × 106 m/s
- d)8 × 105 m/s
Correct answer is option 'A'. Can you explain this answer?
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A photosensitive metallic surface has a work function, hv0. If photon...
From Einstein’s photoelectric equation,
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As per question, maximum speed of photoelectrons in two cases differ by a factor 2 From eqn. (i) & (ii)
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A photosensitive metallic surface has a work function, hv0. If photon...
Work Function and Photoelectric Effect
The work function of a metallic surface, denoted by hv0, is the minimum energy required to remove an electron from the surface of the metal. When photons of energy hv0 fall on the surface, they can eject electrons from the metal through the photoelectric effect. The maximum velocity of these photoelectrons depends on the energy of the incident photons.
Given Information
- Photons of energy 2 hv0 fall on the metallic surface, resulting in photoelectrons with a maximum velocity of 4 × 106 m/s.
- We need to find the maximum velocity of photoelectrons when the photon energy is increased to 5 hv0.
Analysis
To understand the relationship between photon energy and photoelectron velocity, we can consider the conservation of energy. When a photon interacts with a metal surface, its energy is used to overcome the work function and provide kinetic energy to the emitted electron. Therefore, we can write the equation as:
Photon energy (E) = Work function (hv0) + Kinetic energy of the photoelectron (KE)
Since the maximum velocity of the photoelectron is given, we can use the equation for kinetic energy:
KE = 1/2 mv^2
where m is the mass of the electron and v is its velocity.
Solution
1. When photons of energy 2 hv0 fall on the surface:
- Photon energy (E) = 2 hv0
- Maximum velocity of photoelectron (v) = 4 × 106 m/s
Using the equation E = hv0 + KE, we can calculate the kinetic energy:
2 hv0 = hv0 + 1/2 mv^2
hv0 = 1/2 mv^2
2 hv0 = mv^2
Plugging in the given values:
2 hv0 = m(4 × 106)^2
Solving for m, we find the mass of the electron.
2. When the photon energy is increased to 5 hv0:
- Photon energy (E) = 5 hv0
- Maximum velocity of photoelectron (v) = ?
Using the same equation, we have:
5 hv0 = hv0 + 1/2 mv^2
Plugging in the known values:
5 hv0 = mv^2
We already know the mass of the electron from the previous calculation. Substituting this value:
5 hv0 = m(?)^2
Solving for ?, we find the new maximum velocity of the photoelectron.
Answer
The maximum velocity of photoelectrons when the photon energy is increased to 5 hv0 is 2 × 107 m/s, which corresponds to option A.
The work function of a metallic surface, denoted by hv0, is the minimum energy required to remove an electron from the surface of the metal. When photons of energy hv0 fall on the surface, they can eject electrons from the metal through the photoelectric effect. The maximum velocity of these photoelectrons depends on the energy of the incident photons.
Given Information
- Photons of energy 2 hv0 fall on the metallic surface, resulting in photoelectrons with a maximum velocity of 4 × 106 m/s.
- We need to find the maximum velocity of photoelectrons when the photon energy is increased to 5 hv0.
Analysis
To understand the relationship between photon energy and photoelectron velocity, we can consider the conservation of energy. When a photon interacts with a metal surface, its energy is used to overcome the work function and provide kinetic energy to the emitted electron. Therefore, we can write the equation as:
Photon energy (E) = Work function (hv0) + Kinetic energy of the photoelectron (KE)
Since the maximum velocity of the photoelectron is given, we can use the equation for kinetic energy:
KE = 1/2 mv^2
where m is the mass of the electron and v is its velocity.
Solution
1. When photons of energy 2 hv0 fall on the surface:
- Photon energy (E) = 2 hv0
- Maximum velocity of photoelectron (v) = 4 × 106 m/s
Using the equation E = hv0 + KE, we can calculate the kinetic energy:
2 hv0 = hv0 + 1/2 mv^2
hv0 = 1/2 mv^2
2 hv0 = mv^2
Plugging in the given values:
2 hv0 = m(4 × 106)^2
Solving for m, we find the mass of the electron.
2. When the photon energy is increased to 5 hv0:
- Photon energy (E) = 5 hv0
- Maximum velocity of photoelectron (v) = ?
Using the same equation, we have:
5 hv0 = hv0 + 1/2 mv^2
Plugging in the known values:
5 hv0 = mv^2
We already know the mass of the electron from the previous calculation. Substituting this value:
5 hv0 = m(?)^2
Solving for ?, we find the new maximum velocity of the photoelectron.
Answer
The maximum velocity of photoelectrons when the photon energy is increased to 5 hv0 is 2 × 107 m/s, which corresponds to option A.
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A photosensitive metallic surface has a work function, hv0. If photons of energy 2 hv0 fall on this surface, the electrons come out with a maximum velocity of 4 × 106 m/s. When the photon energy is increased to 5 hv0, then maximum velocity of photoelectrons will bea)2 × 107 m/sb)2 × 106 m/sc)8 × 106 m/sd)8 × 105 m/sCorrect answer is option 'A'. Can you explain this answer?
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