A photon of frequency `v` causes photo electric emission from a surface with threshold frequency `v_(0)`. The de-Broglies wavelength `(lambda)` of the photo electron emitted is given by:

Options

A. `lambda=(h)/(2m trianglev)`
B. `lambda=(h)/( trianglev)`
C. `lambda=[(h)/(2m trianglev)]^(1//2)`
D. `lambda=[(h)/(2m trianglev)]^(2)`

Answer

To derive the de Broglie wavelength \((\lambda)\) of the photoelectron emitted during the photoelectric effect, we can follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect occurs when a photon of frequency \(\nu\) strikes a surface and causes the emission of electrons. The energy of the incoming photon must be greater than the work function (threshold energy) of the material for electrons to be emitted.

### Step 2: Write the Energy Equation
The energy of the photon can be expressed as:
\[ E = h\nu \]
where \(h\) is Planck's constant and \(\nu\) is the frequency of the photon.

The work function (threshold energy) is given by:
\[ E_0 = h\nu_0 \]
where \(\nu_0\) is the threshold frequency.

When a photon strikes the surface, the energy can be divided into the work function and the kinetic energy of the emitted electron:
\[ h\nu = h\nu_0 + KE \]
where \(KE\) is the kinetic energy of the emitted electron.

### Step 3: Express Kinetic Energy
Rearranging the equation gives us:
\[ KE = h\nu - h\nu_0 = h(\nu - \nu_0) \]
Let’s denote \(\Delta \nu = \nu - \nu_0\). Thus, we have:
\[ KE = h\Delta \nu \]

### Step 4: Relate Kinetic Energy to Velocity
The kinetic energy of the emitted electron can also be expressed in terms of its mass \(m\) and velocity \(v\):
\[ KE = \frac{1}{2}mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2}mv^2 = h\Delta \nu \]

### Step 5: Solve for Velocity
From the above equation, we can solve for \(v^2\):
\[ v^2 = \frac{2h\Delta \nu}{m} \]

### Step 6: Use de Broglie's Wavelength Formula
According to de Broglie's hypothesis, the wavelength \(\lambda\) of a particle is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting \(v\) from our previous equation:
\[ \lambda = \frac{h}{m\sqrt{\frac{2h\Delta \nu}{m}}} \]
This simplifies to:
\[ \lambda = \frac{h}{\sqrt{2mh\Delta \nu}} \]

### Step 7: Final Expression
Rearranging gives:
\[ \lambda = \sqrt{\frac{h}{2m\Delta \nu}} \]

### Conclusion
Thus, the de Broglie wavelength \((\lambda)\) of the photoelectron emitted is given by:
\[ \lambda = \sqrt{\frac{h}{2m(\nu - \nu_0)}} \]