How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$? (Let $E$ be the initial energy of the electron).

$A$ material particle with a rest mass $m_0$ is moving with the velocity of light $C$. Then,the wavelength of the de$-$Broglie wave associated with it is

Electrons with de-Broglie wavelength $\lambda$ fall on the target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-rays is

$A$ particle of mass $1\, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6\, m\,s^{-1}$. The velocity of the particle is (Mass of electron $= 9.1 \times 10^{-31}\, kg$)

The de Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths is . . . . . . .

If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is