Question:
An electron of mass 'm' and a photon have same energy 'E'. The ratio of de-Broglie wavelength of electron to the wavelength of photon is (c = velocity of light)
An electron of mass 'm' and a photon have same energy 'E'. The ratio of de-Broglie wavelength of electron to the wavelength of photon is (c = velocity of light)
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You can verify the correct option quickly using dimensional analysis. Since a wavelength ratio is a dimensionless number, the right side of the equation must also be dimensionless. The term $\sqrt{\frac{E}{m}}$ has units of velocity ($\text{m s}^{-1}$). Dividing it by the speed of light $c$ ($\text{m s}^{-1}$) is the only way to cancel out all physical units.
Updated On: Jun 12, 2026
- $c\sqrt{\frac{E}{m}}$
- $\frac{1}{c}\sqrt{\frac{2m}{E}}$
- $\frac{1}{c}\sqrt{\frac{E}{2m}}$
- $c\sqrt{\frac{m}{E}}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
An electron (a massive material particle) and a photon (a massless quantum of light) share the exact same energy value $E$. We need to derive the mathematical expression for the ratio of the electron's de-Broglie wavelength ($\lambda_e$) to the photon's electromagnetic wavelength ($\lambda_p$).
Step 2: Key Formula or Approach:
1. The de-Broglie wavelength of a particle with mass $m$ and kinetic energy $E$ is given by:
$$\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$ 2. The wavelength of a photon carrying energy $E$ is derived from the Planck-Einstein relation:
$$E = \frac{hc}{\lambda_p} \implies \lambda_p = \frac{hc}{E}$$ We divide these two wavelength equations to find the required ratio.
Step 3: Detailed Explanation:
Let's write out the ratio of the electron's wavelength to the photon's wavelength:
$$\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}$$ We can simplify this fraction by canceling Planck's constant $h$ from both numerators:
$$\frac{\lambda_e}{\lambda_p} = \frac{1}{\sqrt{2mE}} \times \frac{E}{c} = \frac{E}{c\sqrt{2mE}}$$ Next, bring the energy variable $E$ from the numerator inside the radical sign in the denominator as $E^2$ to simplify the expression:
$$\frac{\lambda_e}{\lambda_p} = \frac{1}{c} \sqrt{\frac{E^2}{2mE}}$$ Cancel a factor of $E$ from the numerator and denominator inside the square root:
$$\frac{\lambda_e}{\lambda_p} = \frac{1}{c} \sqrt{\frac{E}{2m}}$$ This matching derivation corresponds to the expression presented in option (C).
Step 4: Final Answer:
The ratio of their wavelengths is $\frac{1}{c}\sqrt{\frac{E}{2m}}$, which corresponds to option (C).
An electron (a massive material particle) and a photon (a massless quantum of light) share the exact same energy value $E$. We need to derive the mathematical expression for the ratio of the electron's de-Broglie wavelength ($\lambda_e$) to the photon's electromagnetic wavelength ($\lambda_p$).
Step 2: Key Formula or Approach:
1. The de-Broglie wavelength of a particle with mass $m$ and kinetic energy $E$ is given by:
$$\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$$ 2. The wavelength of a photon carrying energy $E$ is derived from the Planck-Einstein relation:
$$E = \frac{hc}{\lambda_p} \implies \lambda_p = \frac{hc}{E}$$ We divide these two wavelength equations to find the required ratio.
Step 3: Detailed Explanation:
Let's write out the ratio of the electron's wavelength to the photon's wavelength:
$$\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}$$ We can simplify this fraction by canceling Planck's constant $h$ from both numerators:
$$\frac{\lambda_e}{\lambda_p} = \frac{1}{\sqrt{2mE}} \times \frac{E}{c} = \frac{E}{c\sqrt{2mE}}$$ Next, bring the energy variable $E$ from the numerator inside the radical sign in the denominator as $E^2$ to simplify the expression:
$$\frac{\lambda_e}{\lambda_p} = \frac{1}{c} \sqrt{\frac{E^2}{2mE}}$$ Cancel a factor of $E$ from the numerator and denominator inside the square root:
$$\frac{\lambda_e}{\lambda_p} = \frac{1}{c} \sqrt{\frac{E}{2m}}$$ This matching derivation corresponds to the expression presented in option (C).
Step 4: Final Answer:
The ratio of their wavelengths is $\frac{1}{c}\sqrt{\frac{E}{2m}}$, which corresponds to option (C).
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