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 solve the problem of finding the de Broglie wavelength of an electron emitted from a filament and accelerated through a voltage \( V \), we can follow these steps:

### Step 1: Understand the de Broglie Wavelength Formula
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

**Hint:** Recall that momentum \( p \) can be expressed in terms of kinetic energy.

### Step 2: Determine the Kinetic Energy of the Electron
When the electron is accelerated through a potential difference \( V \), its kinetic energy \( KE \) can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron.

**Hint:** Remember that the kinetic energy gained by a charged particle when accelerated through a potential difference is equal to the charge times the potential difference.

### Step 3: Relate Kinetic Energy to Momentum
The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the electron. Setting the two expressions for kinetic energy equal gives:
\[ eV = \frac{p^2}{2m} \]

**Hint:** You can rearrange this equation to solve for momentum \( p \).

### Step 4: Solve for Momentum
From the equation \( eV = \frac{p^2}{2m} \), we can solve for momentum \( p \):
\[ p^2 = 2m eV \]
Taking the square root:
\[ p = \sqrt{2m eV} \]

**Hint:** Make sure to keep track of the units for mass and charge when substituting values.

### Step 5: Substitute Momentum into the de Broglie Wavelength Formula
Now substitute \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m eV}} \]

**Hint:** Ensure you know the values of Planck's constant \( h \) and the mass of the electron \( m \).

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

### Summary
The final answer for the de Broglie wavelength of the electron is:
\[ \lambda = \frac{h}{\sqrt{2m eV}} \]