Electron has energy of 100 eV what will be its wavelength &

Options

A. 1.2Å
B. 10Å
C. 100Å
D. 1Å

Answer

To find the wavelength of an electron with an energy of 100 eV, we can use the de Broglie wavelength formula, which relates the wavelength (λ) to the momentum (P) of the particle. The steps to solve the problem are as follows:

### Step 1: Convert Energy from eV to Joules
The energy given is in electron volts (eV), and we need to convert it to joules (J) using the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Thus,
\[ E = 100 \text{ eV} = 100 \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-17} \text{ J} \]

### Step 2: Calculate Momentum (P)
The kinetic energy (E) of the electron is related to its momentum (P) by the formula:
\[ E = \frac{P^2}{2m} \]
Rearranging this gives:
\[ P = \sqrt{2mE} \]
Where:
- \( m \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \text{ kg} \).

Substituting the values:
\[ P = \sqrt{2 \times (9.1 \times 10^{-31} \text{ kg}) \times (1.6 \times 10^{-17} \text{ J})} \]

### Step 3: Calculate the Value of P
Calculating the above expression:
\[ P = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}} \]
\[ P \approx \sqrt{2.912 \times 10^{-47}} \]
\[ P \approx 5.39 \times 10^{-24} \text{ kg m/s} \]

### Step 4: Use de Broglie Wavelength Formula
The de Broglie wavelength (λ) is given by:
\[ \lambda = \frac{h}{P} \]
Where \( h \) (Planck's constant) is approximately \( 6.626 \times 10^{-34} \text{ J s} \).

Substituting the values:
\[ \lambda = \frac{6.626 \times 10^{-34}}{5.39 \times 10^{-24}} \]

### Step 5: Calculate the Wavelength
Calculating the above expression:
\[ \lambda \approx 1.23 \times 10^{-10} \text{ m} \]

### Step 6: Convert to Angstroms
To convert meters to angstroms (1 Å = \( 10^{-10} \) m):
\[ \lambda \approx 1.23 \text{ Å} \]

### Final Answer
The wavelength of the electron with an energy of 100 eV is approximately \( 1.23 \text{ Å} \).

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