Photoelectric Effect + de-Broglie Wavelength Combined Question | JEE Main Concept 🚀

 

💡 Question

A photoemissive substance is illuminated with radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$.
The longest wavelength of radiation that can just emit photoelectrons (i.e., threshold wavelength) is $\lambda_0$.

Find the expression for de-Broglie wavelength ($\lambda_e$) of the emitted electrons in terms of $\lambda_i$ and $\lambda_0$.

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Photoelectric effect equation

When light of wavelength $\lambda_i$ falls on a metal surface, the energy of the incident photon is:

$$E=\frac{hc}{{\lambda }_{\mathrm{i}}}$$

The maximum kinetic energy of the emitted electron is:

$$K_{\text{max}}=\frac{hc}{{\lambda }_i}-\frac{hc}{{\lambda }_{\mathrm{0}}}$$

where $\frac{hc}{\lambda_0}$ is the work function $\phi$ (minimum energy required to eject an electron).

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Step 2 — Relate kinetic energy and de-Broglie wavelength

For an electron having de-Broglie wavelength $\lambda_e$:

$$K_{\text{max}}=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }_e^{\mathrm{2}}}$$

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Step 3 — Equating the two expressions for $K_{\text{max}}$

$$\frac{h^2}{2m{\lambda }_e^2}=hc\text{\hspace{0.17em}⁣}(\frac{1}{{\lambda }_i}-\frac{1}{{\lambda }_0})$$

Simplify to isolate $\lambda_e$:

$${\lambda }_e=\sqrt{\frac{h}{2mc}\cdot \frac{1}{(\frac{1}{{\lambda }_i}-\frac{1}{{\lambda }_0})}}$$

Or, more compactly:

$$\boxed{{\displaystyle {\displaystyle \frac{1}{{\lambda }_e^2}=\frac{2mc}{h}(\frac{1}{{\lambda }_i}-\frac{1}{{\lambda }_0})}}}$$

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🧠 Concept Recap

Concept	Formula	Meaning
Photon Energy	$E = \frac{hc}{\lambda}$	Energy of an incoming photon
Work Function	$\phi = \frac{hc}{\lambda_0}$	Minimum energy to eject an electron
Kinetic Energy of Emitted Electron	$K = E - \phi$	Difference of photon energy & work function
de-Broglie Relation	$\lambda_e = \frac{h}{p}$	Wavelength associated with moving electron

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🖼️ Image Solution

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✅ Final Answer

$$\boxed{{\displaystyle \frac{1}{{\lambda }_e^2}=\frac{2mc}{h}\text{\hspace{0.17em}⁣}(\frac{1}{{\lambda }_i}-\frac{1}{{\lambda }_0})}}$$

or equivalently,

$$\boxed{ \lambda_e = \sqrt{\frac{h}{2mc\,\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}} }$$