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 $eV$)

  • A
    $0.48$
  • B
    $0.68$
  • C
    $0.38$
  • D
    $0.25$

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The de Broglie wavelength for an electron accelerated through a potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is . . . . . . .

If particles move with the same velocity,the de Broglie wavelength is maximum for .........

$A$ particle is moving along the $x-$axis back and forth in a box of length $L$. Assuming the de-Broglie hypothesis is applicable for the particle,and it is moving with a constant speed in the box making perfectly elastic collisions with the walls. The possible value of the momentum of the particle is (Note: $h$ is Planck's constant,$n$ is the principal quantum number).

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2} \, nm$,the potential difference is $......V$.

The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?

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