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Can you explain the answer of this question below:
The threshold frequency v0 for a metal is 7.0 x 1014 s-1. Radiation of frequency v = 1.0 x 1015 s-1 hits the metal. Kinetic energy of the emitted electron is
  • A:
    2.0 x 10-18 J
  • B:
    1.60 x 10-17 J
  • C:
    1.60 x 10-19 J
  • D:
    2.0 x 10-19 J
The answer is d.
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Can you explain the answer of this question below:The threshold freque...
By photoelectric effect



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Can you explain the answer of this question below:The threshold freque...
K.E= h(f-f•).K.E= 6.625 × 10^ -34 (1×10^15- 7×10^14).K.E= 6.625×10^-34 (10-7) × 10^14.K.E= 6.625×3 × 10^-20.k.E= 19.8× 10^-20.as 19.8 is approximately 20....K .E= 20×10^-20.K.E= 2×10^-19 j.
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Can you explain the answer of this question below:The threshold freque...
Threshold Frequency and the Photoelectric Effect

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to electromagnetic radiation, such as light. The energy of the emitted electrons depends on the frequency of the incident radiation and the specific material involved.

Threshold frequency, denoted as v0, is the minimum frequency of radiation required to liberate electrons from the material. If the frequency of the incident radiation is below the threshold frequency, no electrons will be emitted regardless of the intensity of the light. However, if the frequency of the incident radiation is equal to or greater than the threshold frequency, electrons will be emitted.

Given Information:

Threshold frequency (v0) = 7.0 x 10^14 s^-1
Frequency of incident radiation (v) = 1.0 x 10^15 s^-1

Determining the Kinetic Energy of the Emitted Electron

To determine the kinetic energy of the emitted electron, we need to compare the frequency of the incident radiation with the threshold frequency. If the frequency of the incident radiation is greater than or equal to the threshold frequency, electrons will be emitted and will possess kinetic energy.

Comparing the Frequencies

The frequency of the incident radiation (v = 1.0 x 10^15 s^-1) is greater than the threshold frequency (v0 = 7.0 x 10^14 s^-1). Therefore, electrons will be emitted from the metal in question.

Calculating the Kinetic Energy

The kinetic energy (KE) of an electron emitted in the photoelectric effect can be calculated using the equation:

KE = hv - hv0

where h is the Planck's constant (6.63 x 10^-34 J·s), v is the frequency of the incident radiation, and v0 is the threshold frequency.

Substituting the given values into the equation:

KE = (6.63 x 10^-34 J·s)(1.0 x 10^15 s^-1) - (6.63 x 10^-34 J·s)(7.0 x 10^14 s^-1)

Simplifying the equation:

KE = 6.63 x 10^-34 J - 46.41 x 10^-34 J

KE = -39.78 x 10^-34 J

The negative sign indicates that the emitted electron has a negative kinetic energy, which is not physically possible. Therefore, the kinetic energy of the emitted electron is zero.

Answer: The kinetic energy of the emitted electron is zero.
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Can you explain the answer of this question below:The threshold frequency v0 for a metal is 7.0 x 1014 s-1. Radiation of frequency v = 1.0 x 1015 s-1 hits the metal. Kinetic energy of the emitted electron isA:2.0 x 10-18 JB:1.60 x 10-17 JC:1.60 x 10-19 JD:2.0 x 10-19 JThe answer is d.
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