A photon of frequency n causes photoelectric emmission from a surface with threshold frequency `n_0` .The de Broglie wavelength `lambda ` of the photoelectron emitted is given as

Options

A. `Delta n = (h)/(2 m lambda)`
B. `Delta n = (h)/(lambda)`
C. `[(1)/(v_(0)) - (1)/(v)] = (mc^(2))/n`
D. `lambda=sqrt(h/(2mDeltaV))`

Explanation

To solve the problem of finding the de Broglie wavelength \( \lambda \) of the photoelectron emitted due to photoelectric emission, we can follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect occurs when photons strike a material and emit electrons. The energy of the incoming photon must be greater than the work function (or threshold energy) of the material for the emission to occur.

### Step 2: Relate Photon Energy to Frequency
The energy \( E \) of a photon is given by the equation:
\[
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