Logo
Jeeify
MCQEasyJEE 2025Photoelectric Effect

JEE Physics 2025 Question with Solution

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV2.14 \, \text{eV} and the stopping potential is 2 V2 \, \text{V}, what is the wavelength of the electromagnetic wave?

Given hc=1242 eV⋅nmhc = 1242 \, \text{eV} \cdot \text{nm} where hh is the Planck constant and cc is the speed of light in vacuum.

  • A

    400 nm400 \, \text{nm}

  • B

    600 nm600 \, \text{nm}

  • C

    200 nm200 \, \text{nm}

  • D

    300 nm300 \, \text{nm}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

  • Work function ϕ=2.14 eV\phi = 2.14 \, \text{eV}
  • Stopping potential V0=2 VV_0 = 2 \, \text{V}
  • hc=1242 eV⋅nmhc = 1242 \, \text{eV} \cdot \text{nm}

Find: The wavelength λ\lambda of the incident electromagnetic wave.

In the photoelectric effect, the photon energy is used to overcome the work function and provide the maximum kinetic energy of the emitted electrons.

E=ϕ+eV0=2.14 eV+2 eV=4.14 eVE = \phi + eV_0 = 2.14 \, \text{eV} + 2 \, \text{eV} = 4.14 \, \text{eV}

Also,

E=hcλE = \frac{hc}{\lambda}

Substituting the given values,

4.14=1242λ4.14 = \frac{1242}{\lambda}

Therefore,

λ=12424.14≈300 nm\lambda = \frac{1242}{4.14} \approx 300 \, \text{nm}

Therefore, the wavelength of the electromagnetic wave is 300 nm300 \, \text{nm}. The correct option is D.

Direct Energy Balance

Given: ϕ=2.14 eV\phi = 2.14 \, \text{eV}, V0=2 VV_0 = 2 \, \text{V}, and hc=1242 eV⋅nmhc = 1242 \, \text{eV} \cdot \text{nm}.

Find: λ\lambda.

The stopping potential of 2 V2 \, \text{V} means the maximum kinetic energy is 2 eV2 \, \text{eV}. So the photon energy is immediately

Ephoton=2.14+2=4.14 eVE_{\text{photon}} = 2.14 + 2 = 4.14 \, \text{eV}

Now use

λ=hcE=12424.14≈300 nm\lambda = \frac{hc}{E} = \frac{1242}{4.14} \approx 300 \, \text{nm}

This works because, in electron-volt units, eV0eV_0 is numerically equal to the stopping potential in volts for a single electron. Hence the correct option is D.

Common mistakes

  • Using only the work function and ignoring the stopping potential is incorrect because the photon energy must supply both the work function and the maximum kinetic energy. Instead, use E=ϕ+eV0E = \phi + eV_0.

  • Treating the stopping potential 2 V2 \, \text{V} as an energy of 2 J2 \, \text{J} is wrong. For a single electron, the corresponding kinetic energy is 2 eV2 \, \text{eV}, not joules.

  • Reversing the wavelength relation and calculating λ=Ehc\lambda = \frac{E}{hc} is incorrect. The correct relation is E=hcλE = \frac{hc}{\lambda}, so λ=hcE\lambda = \frac{hc}{E}.

Practice more Photoelectric Effect questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Photoelectric Effect

Related questions