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The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons is
- a)0.75V
- b)1.75V
- c)2.5V
- d)3.75V
Correct answer is option 'B'. Can you explain this answer?
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To determine the stopping potential of photoelectrons emitted when Potassium is illuminated by light of wavelength 3300 A0, we need to use the concept of the photoelectric effect and the equation for the stopping potential.
The photoelectric effect states that when light of a certain frequency or wavelength is incident on a metal surface, electrons are emitted from the surface. The energy of the incident photons must be greater than or equal to the work function of the metal for this emission to occur.
Given data:
- Work function of Potassium (φ) = 2.0 eV
- Wavelength of incident light (λ) = 3300 A0 (1 A0 = 10^-10 m)
1. Calculate the energy of the incident photons:
The energy (E) of a photon can be calculated using the equation:
E = hc/λ
where h is the Planck's constant (6.63 x 10^-34 J.s) and c is the speed of light (3 x 10^8 m/s).
E = (6.63 x 10^-34 J.s * 3 x 10^8 m/s) / (3300 x 10^-10 m)
E ≈ 6.02 x 10^-19 J
2. Convert the energy to electron volts (eV):
1 eV = 1.6 x 10^-19 J
E = (6.02 x 10^-19 J) / (1.6 x 10^-19 J/eV)
E ≈ 3.76 eV
3. Calculate the stopping potential (V):
The stopping potential (V) is the minimum potential difference required to stop the emitted photoelectrons from reaching the anode.
The stopping potential can be calculated using the equation:
eV = E - φ
where e is the elementary charge (1.6 x 10^-19 C).
eV = (3.76 eV) - (2.0 eV)
V ≈ 1.76 V
Therefore, the stopping potential of the photoelectrons emitted when Potassium is illuminated by light of wavelength 3300 A0 is approximately 1.76 V.
Option 'B' (1.75V) is the closest value to the calculated stopping potential.
The photoelectric effect states that when light of a certain frequency or wavelength is incident on a metal surface, electrons are emitted from the surface. The energy of the incident photons must be greater than or equal to the work function of the metal for this emission to occur.
Given data:
- Work function of Potassium (φ) = 2.0 eV
- Wavelength of incident light (λ) = 3300 A0 (1 A0 = 10^-10 m)
1. Calculate the energy of the incident photons:
The energy (E) of a photon can be calculated using the equation:
E = hc/λ
where h is the Planck's constant (6.63 x 10^-34 J.s) and c is the speed of light (3 x 10^8 m/s).
E = (6.63 x 10^-34 J.s * 3 x 10^8 m/s) / (3300 x 10^-10 m)
E ≈ 6.02 x 10^-19 J
2. Convert the energy to electron volts (eV):
1 eV = 1.6 x 10^-19 J
E = (6.02 x 10^-19 J) / (1.6 x 10^-19 J/eV)
E ≈ 3.76 eV
3. Calculate the stopping potential (V):
The stopping potential (V) is the minimum potential difference required to stop the emitted photoelectrons from reaching the anode.
The stopping potential can be calculated using the equation:
eV = E - φ
where e is the elementary charge (1.6 x 10^-19 C).
eV = (3.76 eV) - (2.0 eV)
V ≈ 1.76 V
Therefore, the stopping potential of the photoelectrons emitted when Potassium is illuminated by light of wavelength 3300 A0 is approximately 1.76 V.
Option 'B' (1.75V) is the closest value to the calculated stopping potential.
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The work function of Potassium is 2.0 eV. When it is illuminated by light of wavelength 3300 A0 photoelectrons are emitted. The stopping potential of photoelectrons isa)0.75Vb)1.75Vc)2.5Vd)3.75VCorrect answer is option 'B'. Can you explain this answer?
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