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The log - log graph between the energy E...

The log - log graph between the energy `E` of an electron and its de - Broglie wavelength `lambda` will be

A

B

C

D

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To analyze the relationship between the energy \( E \) of an electron and its de Broglie wavelength \( \lambda \), we can start from the de Broglie wavelength formula and derive the log-log relationship step by step. ### Step-by-Step Solution: 1. **Understanding the de Broglie Wavelength**: The de Broglie wavelength \( \lambda \) of a particle is given by the formula: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. 2. **Relating Momentum to Energy**: For an electron, the momentum \( p \) can be expressed in terms of its kinetic energy \( E \): \[ p = \sqrt{2mE} \] where \( m \) is the mass of the electron. 3. **Substituting Momentum into the Wavelength Formula**: By substituting the expression for momentum into the de Broglie wavelength formula, we get: \[ \lambda = \frac{h}{\sqrt{2mE}} \] 4. **Taking the Logarithm of Both Sides**: To analyze the relationship in a log-log graph, we take the logarithm of both sides: \[ \log(\lambda) = \log\left(\frac{h}{\sqrt{2mE}}\right) \] 5. **Applying Logarithmic Properties**: Using the properties of logarithms, we can separate the terms: \[ \log(\lambda) = \log(h) - \log(\sqrt{2m}) - \log(\sqrt{E}) \] This simplifies to: \[ \log(\lambda) = \log(h) - \frac{1}{2}\log(2m) - \frac{1}{2}\log(E) \] 6. **Rearranging the Equation**: Rearranging gives us: \[ \log(\lambda) = \log(h) - \frac{1}{2}\log(2m) - \frac{1}{2}\log(E) \] This can be expressed as: \[ \log(\lambda) = -\frac{1}{2}\log(E) + \left(\log(h) - \frac{1}{2}\log(2m)\right) \] 7. **Identifying the Log-Log Graph**: In the equation \( \log(\lambda) = -\frac{1}{2}\log(E) + C \), where \( C = \log(h) - \frac{1}{2}\log(2m) \), we can see that: - The slope of the graph (in a log-log plot) is \( -\frac{1}{2} \). - The intercept on the \( y \)-axis is \( C \). ### Conclusion: The log-log graph between the energy \( E \) of an electron and its de Broglie wavelength \( \lambda \) will have a slope of \( -\frac{1}{2} \) and a positive intercept on the \( y \)-axis.

To analyze the relationship between the energy \( E \) of an electron and its de Broglie wavelength \( \lambda \), we can start from the de Broglie wavelength formula and derive the log-log relationship step by step. ### Step-by-Step Solution: 1. **Understanding the de Broglie Wavelength**: The de Broglie wavelength \( \lambda \) of a particle is given by the formula: \[ \lambda = \frac{h}{p} ...
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Knowledge Check

  • de-Broglie wavelength lambda is

    A
    proportional to mass
    B
    proportional to impulse
    C
    inversely proportional to impulse
    D
    independent to impulse
  • The de Broglie wavelength lambda of a particle

    A
    is proportional to mass
    B
    is proportional to impluse
    C
    is inversely proportional to impulse
    D
    does not depend on impulse
  • The de Broglie wavelength lambda of a particle

    A
    is proportional to mass
    B
    is proportional to impluse
    C
    is inversely proportional to impulse
    D
    does not depend on impulse
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