What is the de-Broglie wavelength of a bullet of mass $0.033$ kg travelling at the speed of $1$ km/s? $(h = 6.6 \times 10^{-34} \text{ Js})$

  • A
    $3 \times 10^{-25} \text{ m}$
  • B
    $2 \times 10^{-35} \text{ m}$
  • C
    $1.1 \times 10^{-32} \text{ m}$
  • D
    $1.7 \times 10^{-35} \text{ m}$

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If the de Broglie wavelength of an electron is $2 \ nm$,then its kinetic energy is nearly (Planck's constant $= 6.6 \times 10^{-34} \ J \ s$ and mass of electron $= 9 \times 10^{-31} \ kg$) (in $eV$)

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with electrons change?

Two particles move at right angles to each other. Their de Broglie wavelengths are $\lambda_1$ and $\lambda_2$ respectively. The particles undergo a perfectly inelastic collision. The de Broglie wavelength $\lambda$ of the final particle is given by

If we consider an electron and a photon of the same de-Broglie wavelength,then they will have the same:

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is ($c$ = velocity of light in air).

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