The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \mathrm{~m} /…

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$ is equal to that of a photon. The ratio of kinetic energy of the electron to that of the photon $\left(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$
  1. $2$
  2. $4$
  3. $\frac{1}{2}$
  4. $\frac{1}{4}$

Solution

The ratio of kinetic energy of the electron to that of the photon. $\begin{aligned} & =\frac{v}{2 c} \\ & =\frac{1.5 \times 10^8}{2 \times 3 \times 10^8}=\frac{1}{4}\end{aligned}$

Asked in: AP EAMCET 2012

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