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De Broglie Wavelength from Photoelectric Effect Formula

Learn how to derive the de Broglie wavelength of emitted electrons using incident and threshold wavelengths. This combines photoelectric effect and...

 

đź’ˇ Question

A photoemissive substance is illuminated with radiation of wavelength λi\lambda_i so that it releases electrons with de-Broglie wavelength λe\lambda_e.
The longest wavelength of radiation that can just emit photoelectrons (i.e., threshold wavelength) is λ0\lambda_0.

Find the expression for de-Broglie wavelength (λe\lambda_e) of the emitted electrons in terms of λi\lambda_i and λ0\lambda_0.


đź–Ľ️ Question Image

A photoemissive substance is illuminated with a radiation of wavelength λᵢ so that it releases electrons with de-Broglie wavelength λₑ. The longest wavelength of radiation that can emit photoelectron is λ₀. Expression for de-Broglie wavelength is given by:


✍️ Short Explanation

This problem is based on:

👉 Photoelectric effect
👉 Threshold wavelength
👉 de-Broglie wavelength.

Main idea:

Use Einstein’s photoelectric equation:

hcλi=hcλ0+K\boxed{ \frac{hc}{\lambda_i} = \frac{hc}{\lambda_0}+K }

and:

λe=hp\boxed{ \lambda_e=\frac{h}{p} }


De Broglie Wavelength from Photoelectric Effect Formula


đź”· Step 1 — Write Photoelectric Equation đź’Ż

Photon energy:

hcλi\frac{hc}{\lambda_i}

Threshold energy:

hcλ0\frac{hc}{\lambda_0}

Thus kinetic energy of emitted electron:

K=hcλi−hcλ0K= \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0}
K=hc(1λi−1λ0)\boxed{ K=hc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) }


đź”· Step 2 — Relate KE with Momentum

For electron:

K=p22mK=\frac{p^2}{2m}

Thus:

p22m=hc(1λi−1λ0)\frac{p^2}{2m} = hc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right)

So:

p=2mhc(1λi−1λ0)p= \sqrt{ 2mhc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) }


đź”· Step 3 — Apply de-Broglie Relation

λe=hp\lambda_e=\frac{h}{p}

Substitute pp:

λe=h2mhc(1λi−1λ0)\lambda_e = \frac{h}{ \sqrt{ 2mhc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) }}

Take hh inside square root:

λe=h2mc(1λi−1λ0)\boxed{ \lambda_e = \sqrt{ \frac{ h }{ 2mc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) } } }


đź”· Step 4 — JEE Trap Alert 🚨

❌ Threshold wavelength and incident wavelength confuse kar dena

❌ de-Broglie formula directly energy se relate na kar pana

❌ Kinetic energy sign mistake

Remember:

K=Eincident−Ď•\boxed{ K=E_{\text{incident}}-\phi }


✅ Final Answer

λe=h2mc(1λi−1λ0)

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