The Davisson-Germer experiments that first demonstrated the wave nature of matter used electron accelerated to `54eV`. The wavelength (in `A^o` ) of the electrons in the Davisson-Germer experiment is

Answer

To find the wavelength of electrons accelerated to 54 eV in the Davisson-Germer experiment, we can use the de Broglie wavelength formula and the kinetic energy of the electrons. Here’s a step-by-step solution:

### Step 1: Understand the relationship between kinetic energy and velocity
The kinetic energy (K.E.) of an electron can be expressed as:
\[ K.E. = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the electron and \(v\) is its velocity.

### Step 2: Relate kinetic energy to the energy in electron volts
The kinetic energy given is 54 eV. To convert this energy into joules, we use the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\):
\[ K.E. = 54 \text{ eV} = 54 \times 1.6 \times 10^{-19} \text{ J} \]

### Step 3: Calculate the kinetic energy in joules
Calculating the kinetic energy:
\[ K.E. = 54 \times 1.6 \times 10^{-19} = 8.64 \times 10^{-18} \text{ J} \]

### Step 4: Solve for velocity using kinetic energy
From the kinetic energy formula, we can solve for \(v\):
\[ v = \sqrt{\frac{2 \times K.E.}{m}} \]
Substituting the values, where the mass of the electron \(m = 9.1 \times 10^{-31} \text{ kg}\):
\[ v = \sqrt{\frac{2 \times 8.64 \times 10^{-18}}{9.1 \times 10^{-31}}} \]

### Step 5: Calculate the velocity
Calculating \(v\):
\[ v = \sqrt{\frac{1.728 \times 10^{-17}}{9.1 \times 10^{-31}}} \approx \sqrt{1.8978 \times 10^{13}} \approx 4.36 \times 10^6 \text{ m/s} \]

### Step 6: Use de Broglie wavelength formula
The de Broglie wavelength \(\lambda\) is given by:
\[ \lambda = \frac{h}{mv} \]
where \(h\) is Planck's constant, \(h = 6.63 \times 10^{-34} \text{ J s}\).

### Step 7: Substitute values to find the wavelength
Substituting the values into the de Broglie wavelength formula:
\[ \lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 4.36 \times 10^6} \]

### Step 8: Calculate the wavelength
Calculating \(\lambda\):
\[ \lambda = \frac{6.63 \times 10^{-34}}{3.96 \times 10^{-24}} \approx 1.67 \times 10^{-10} \text{ m} \]

### Step 9: Convert to angstroms
To convert meters to angstroms (1 Å = \(10^{-10}\) m):
\[ \lambda \approx 1.67 \text{ Å} \]

### Final Answer
The wavelength of the electrons in the Davisson-Germer experiment is approximately **1.67 Å**.

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