De Broglie Wavelength from Photoelectric Effect Formula

 

💡 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 Explanation

This problem is based on:

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

Main idea:

Use Einstein’s photoelectric equation:

$$\boxed{ \frac{hc}{\lambda_i} = \frac{hc}{\lambda_0}+K }$$

and:

$$\boxed{ \lambda_e=\frac{h}{p} }$$

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🔷 Step 1 — Write Photoelectric Equation 💯

Photon energy:

$$\frac{hc}{\lambda_i}$$

Threshold energy:

$$\frac{hc}{\lambda_0}$$

Thus kinetic energy of emitted electron:

$$K= \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0}$$
$$\boxed{ K=hc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) }$$

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🔷 Step 2 — Relate KE with Momentum

For electron:

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

Thus:

$$\frac{p^2}{2m} = hc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right)$$

So:

$$p= \sqrt{ 2mhc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) }$$

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🔷 Step 3 — Apply de-Broglie Relation

$$\lambda_e=\frac{h}{p}$$

Substitute $p$:

$$\lambda_e = \frac{h}{ \sqrt{ 2mhc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) }}$$

Take $h$ inside square root:

$$\boxed{ \lambda_e = \sqrt{ \frac{ h }{ 2mc\left( \frac1{\lambda_i} -\frac1{\lambda_0} \right) } } }$$

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🔷 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:

$$\boxed{ K=E_{\text{incident}}-\phi }$$

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✅ Final Answer

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

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