When energy is added to an electron,its de Broglie wavelength decreases from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$. The added energy is:

Electrons used in an electron microscope are accelerated by a voltage of $25 \; kV$. If the voltage is increased to $100 \; kV$,then the de-Broglie wavelength associated with the electrons would

$A$ proton is accelerated through $225 \,V$. Its de Broglie wavelength is ........ $nm$.

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

An electron accelerated by a potential difference $V$ has a de-Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $9V$,its de-Broglie wavelength will be

The temperature of an ideal gas in $3$-dimensions is $300\, K$. The corresponding de-Broglie wavelength of the electron approximately at $300\, K$ is $....\, nm$.
$[m_e = \text{mass of electron} = 9 \times 10^{-31}\, kg, h = \text{Planck constant} = 6.6 \times 10^{-34}\, Js, k_B = \text{Boltzmann constant} = 1.38 \times 10^{-23}\, JK^{-1}]$