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Mechanics

A radiation of wavelength 'λ' is incident on metal surface having 620 nm threshold wavelength. The stopping potential of electron emitted is 4.9 V. What is the value of λ (in nm)

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7 Years agoGrade
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Profile image of Rituraj Tiwari
5 Years ago

Sure, let's break this down step by step. The key concept here is the photoelectric effect, which is when light (or radiation) of a certain wavelength falls on a metal surface and causes the emission of electrons.

Understanding the Photoelectric Effect

When light of sufficient energy (or wavelength) hits a metal surface, it can knock electrons loose, creating a flow of current. The minimum energy required to release an electron is called the threshold energy, which corresponds to a threshold wavelength.

Finding the Wavelength of the Incident Radiation

In this case, the threshold wavelength of the metal surface is 620 nm. The stopping potential of the emitted electrons is 4.9 V. To find the wavelength of the incident radiation, we can use the formula:

Energy of incident radiation = Work Function + Kinetic Energy

Energy = hc/λ = eV0 + eVstop

Where:

  • h = Planck's constant (6.626 x 10-34 J·s)
  • c = speed of light (3.00 x 108 m/s)
  • λ = wavelength of the incident radiation
  • e = elementary charge (1.6 x 10-19 C)
  • V0 = stopping potential (4.9 V)

Solving for λ

Plugging in the values, we get:

hc/λ = eV0 + eVstop

(6.626 x 10-34 J·s)(3.00 x 108 m/s) / λ = (1.6 x 10-19 C)(4.9 V) + (1.6 x 10-19 C)(4.9 V)

Solving for λ, we find:

λ = (6.626 x 10-34 J·s)(3.00 x 108 m/s) / [(1.6 x 10-19 C)(4.9 V) + (1.6 x 10-19 C)(4.9 V)]

After calculating this, you will get the value of λ in nanometers. This is how you can determine the wavelength of the incident radiation based on the given threshold wavelength and stopping potential.