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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Photoelectric Effect and Wave–Particle Duality

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Photoelectric Effect and Wave–Particle Duality

A photographic film is coated with a layer of silver bromide. When...

A photographic film is coated with a layer of silver bromide. When light strikes this film, the silver bromide molecules dissociate, and the film records the light. A minimum of 0.6 eV0.6 \text{ eV} is required to dissociate a silver bromide molecule. Determine the maximum wavelength of light that can be recorded by the film.

Options

  1. A4140 nm4140 \text{ nm}
  2. B2070 nm2070 \text{ nm}
  3. C3105 nm3105 \text{ nm}
  4. D1035 nm1035 \text{ nm}

Correct answer

B. 2070 nm2070 \text{ nm}

Step-by-step solution

The minimum energy required to dissociate a silver bromide molecule is given as E=0.6 eVE = 0.6 \text{ eV}.

The energy of a photon is related to its wavelength by the equation E=hcλE = \dfrac{hc}{\lambda}.

For maximum wavelength, the energy of the photon must be equal to the minimum dissociation energy.

\lambda_\max = \dfrac{hc}{E_\min}

Substituting hc=1242 eV⋅nmhc = 1242 \text{ eV}\cdot\text{nm} and E_\min = 0.6 \text{ eV}:

\lambda_\max = \dfrac{1242}{0.6} \text{ nm}

\lambda_\max = 2070 \text{ nm}
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