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If a beam of electrons impinges on an energy barrier of height 0.03 eV and of infinite width. Find the fraction of electrons reflected and transmitted at the barrier if the impinging electrons
are having energy 0.04 eV.
are having energy 0.04 eV.
- a)0.111
- b)0.142
- c)0.123
- d)0.175
Correct answer is option 'A'. Can you explain this answer?
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If a beam of electrons impinges on an energy barrier of height 0.03 eV...
Explanation:
The probability of reflection and transmission of electrons at an energy barrier can be calculated using the following formula:
\[ R = \left( \frac{k_2 - k_1}{k_2 + k_1} \right)^2 \]
\[ T = 1 - R \]
Where:
- \( R \) is the fraction of electrons reflected
- \( T \) is the fraction of electrons transmitted
- \( k_1 \) is the wave vector of the incident electrons
- \( k_2 \) is the wave vector of the transmitted electrons
Given:
- Barrier height (\( V \)) = 0.03 eV
- Incident electron energy (\( E \)) = 0.04 eV
First, calculate the wave vectors:
\[ k_1 = \frac{\sqrt{2mE}}{\hbar} \]
\[ k_2 = \frac{\sqrt{2m(E - V)}}{\hbar} \]
Substitute the values:
\[ k_1 = \frac{\sqrt{2 \times 9.11 \times 10^{-31} \times 0.04}}{1.05 \times 10^{-34}} \]
\[ k_1 \approx 2.07 \times 10^9 \]
\[ k_2 = \frac{\sqrt{2 \times 9.11 \times 10^{-31} \times (0.04 - 0.03)}}{1.05 \times 10^{-34}} \]
\[ k_2 \approx 1.46 \times 10^9 \]
Now, calculate the fractions:
\[ R = \left( \frac{1.46 - 2.07}{1.46 + 2.07} \right)^2 \]
\[ R \approx 0.111 \]
\[ T = 1 - 0.111 \]
\[ T \approx 0.889 \]
Therefore, the fraction of electrons reflected at the barrier is 0.111.
The probability of reflection and transmission of electrons at an energy barrier can be calculated using the following formula:
\[ R = \left( \frac{k_2 - k_1}{k_2 + k_1} \right)^2 \]
\[ T = 1 - R \]
Where:
- \( R \) is the fraction of electrons reflected
- \( T \) is the fraction of electrons transmitted
- \( k_1 \) is the wave vector of the incident electrons
- \( k_2 \) is the wave vector of the transmitted electrons
Given:
- Barrier height (\( V \)) = 0.03 eV
- Incident electron energy (\( E \)) = 0.04 eV
First, calculate the wave vectors:
\[ k_1 = \frac{\sqrt{2mE}}{\hbar} \]
\[ k_2 = \frac{\sqrt{2m(E - V)}}{\hbar} \]
Substitute the values:
\[ k_1 = \frac{\sqrt{2 \times 9.11 \times 10^{-31} \times 0.04}}{1.05 \times 10^{-34}} \]
\[ k_1 \approx 2.07 \times 10^9 \]
\[ k_2 = \frac{\sqrt{2 \times 9.11 \times 10^{-31} \times (0.04 - 0.03)}}{1.05 \times 10^{-34}} \]
\[ k_2 \approx 1.46 \times 10^9 \]
Now, calculate the fractions:
\[ R = \left( \frac{1.46 - 2.07}{1.46 + 2.07} \right)^2 \]
\[ R \approx 0.111 \]
\[ T = 1 - 0.111 \]
\[ T \approx 0.889 \]
Therefore, the fraction of electrons reflected at the barrier is 0.111.
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