In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage V then de-Broglie wavelength of that electron will be _______m.

Options

A. `(2 V_(em))/(sqrt(h))`
B. `(sqrt(h))/(2V_(em))`
C. `(sqrt(2V_(em)))/(h)`
D. `(h)/(sqrt(2V_(em)))`

Answer

To find the de Broglie wavelength of an electron that has been accelerated through a voltage \( V \) in the Davisson-Germer experiment, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and voltage
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. The kinetic energy \( KE \) of the electron can be expressed as:
\[ KE = qV \]
where \( q \) is the charge of the electron. For an electron, \( q = e \) (the elementary charge).

### Step 2: Express kinetic energy in terms of momentum
The kinetic energy can also be expressed in terms of momentum \( p \) and mass \( m \) of the electron:
\[ KE = \frac{p^2}{2m} \]
Setting the two expressions for kinetic energy equal gives:
\[ qV = \frac{p^2}{2m} \]

### Step 3: Solve for momentum
Rearranging the equation gives:
\[ p^2 = 2mqV \]
Taking the square root of both sides, we find the momentum \( p \):
\[ p = \sqrt{2mqV} \]

### Step 4: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 5: Substitute momentum into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula, we have:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

### Step 6: Substitute the charge of the electron
Since we are dealing with an electron, we substitute \( q \) with \( e \):
\[ \lambda = \frac{h}{\sqrt{2meV}} \]

### Final Expression
Thus, the de Broglie wavelength of the electron accelerated through voltage \( V \) is:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]