JEE Challenger
More from Dual Nature of Radiation and Matter
PhyMedAdv 2016 • Paper 2

De Broglie Wavelength of Photoelectrons Passing Through Anode

Light having wavelength λph\lambda_{\text{ph}} strikes a cathode plate enclosed inside an evacuated tube as depicted in the diagram. The work function associated with the cathode surface is ϕ\phi, while the anode consists of a conducting wire mesh positioned at a separation dd from the cathode. A potential difference VV is set up across the electrodes. Provided that λe\lambda_e denotes the minimum de Broglie wavelength of electrons crossing through the anode mesh, choose the correct statement(s):

Question Diagram 1

Options

A

λe\lambda_e decreases with increase in ϕ\phi and λph\lambda_{\text{ph}}

B

λe\lambda_e is approximately halved, if dd is doubled

C

For large potential difference (V≫ϕ/eV \gg \phi/e), λe\lambda_e is approximately halved if VV is made four times

Correct
D

λe\lambda_e increases at the same rate as λph\lambda_{\text{ph}} for λph<hc/ϕ\lambda_{\text{ph}} < hc/\phi

Topics & Concepts

Dual Nature of Radiation and MatterPhotoelectric EffectDe Broglie Wavelength

Step-by-Step Solution

To determine the correct statement(s), we begin by applying the photoelectric equation and the concept of de Broglie wavelength for the emitted photoelectrons.

1. Maximum Kinetic Energy of Photoelectrons

According to Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons emitted at the cathode is given by: Kmax, cathode=hcλph−ϕK_{\text{max, cathode}} = \frac{hc}{\lambda_{\text{ph}}} - \phi

As these photoelectrons travel from the cathode to the anode, they are accelerated through a potential difference VV. The energy gained by an electron of charge ee in this potential difference is eVeV.

Therefore, the maximum kinetic energy of the electrons reaching the anode mesh is: Kmax, anode=Kmax, cathode+eV=hcλph−ϕ+eVK_{\text{max, anode}} = K_{\text{max, cathode}} + eV = \frac{hc}{\lambda_{\text{ph}}} - \phi + eV

2. Minimum de Broglie Wavelength

concentrated at the maximum kinetic energy, the minimum de Broglie wavelength λe\lambda_e corresponds to the electrons possessing the maximum momentum pmaxp_{\text{max}} at the anode: λe=hpmax=h2mKmax, anode\lambda_e = \frac{h}{p_{\text{max}}} = \frac{h}{\sqrt{2m K_{\text{max, anode}}}}

Substituting Kmax, anodeK_{\text{max, anode}} into the wavelength expression gives: λe=h2m(hcλph−ϕ+eV)\lambda_e = \frac{h}{\sqrt{2m \left( \frac{hc}{\lambda_{\text{ph}}} - \phi + eV \right)}}

3. Evaluation of Options

  • Option (A): If ϕ\phi or λph\lambda_{\text{ph}} increases, the term (hcλph−ϕ+eV)\left(\frac{hc}{\lambda_{\text{ph}}} - \phi + eV\right) decreases, which means Kmax, anodeK_{\text{max, anode}} decreases. Consequently, λe\lambda_e increases rather than decreases. Thus, Option A is incorrect.

  • Option (B): The expression for λe\lambda_e depends only on the potential difference VV, the work function ϕ\phi, and the photon wavelength λph\lambda_{\text{ph}}. It is completely independent of the distance dd between the cathode and anode. Thus, changing dd has no effect on λe\lambda_e. Thus, Option B is incorrect.

  • Option (C): For a large potential difference such that eV≫ϕeV \gg \phi and eV≫hcλpheV \gg \frac{hc}{\lambda_{\text{ph}}}, the kinetic energy contribution from the potential difference dominates: Kmax, anode≈eVK_{\text{max, anode}} \approx eV

    Under this approximation, the minimum de Broglie wavelength simplifies to: λe≈h2meV\lambda_e \approx \frac{h}{\sqrt{2m eV}}

    If the potential difference is increased to V′=4VV' = 4V, the new wavelength λe′\lambda_e' becomes: λe′≈h2me(4V)=12(h2meV)=λe2\lambda_e' \approx \frac{h}{\sqrt{2m e (4V)}} = \frac{1}{2} \left( \frac{h}{\sqrt{2m eV}} \right) = \frac{\lambda_e}{2}

    Therefore, λe\lambda_e is approximately halved when VV is quadrupled. Thus, Option C is correct.

  • Option (D): Differentiating λe\lambda_e with respect to λph\lambda_{\text{ph}} yields: dλedλph=h2c22m λph2(hcλph−ϕ+eV)3/2\frac{d\lambda_e}{d\lambda_{\text{ph}}} = \frac{h^2 c}{2\sqrt{2m} \, \lambda_{\text{ph}}^2 \left( \frac{hc}{\lambda_{\text{ph}}} - \phi + eV \right)^{3/2}}

    Since dλedλph≠1\frac{d\lambda_e}{d\lambda_{\text{ph}}} \neq 1, λe\lambda_e does not increase at the same rate as λph\lambda_{\text{ph}}. Thus, Option D is incorrect.

Conclusion

The only correct statement is C.

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