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:

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

The de Broglie wavelengths of an electron $(e)$,proton $(p)$,neutron $(n)$,and $\alpha$-particle are $\lambda_e, \lambda_p, \lambda_n$,and $\lambda_\alpha$ respectively. All have the same kinetic energy of $1 \ MeV$. Which of the following represents the correct increasing order of their wavelengths?

If the kinetic energy of an electron and a proton are equal,find the ratio of their associated de Broglie wavelengths.

The de Broglie wavelength of an oxygen molecule at $27^{\circ} C$ is $x \times 10^{-12} \ m$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34} \ J \cdot s$,Boltzmann constant $= 1.38 \times 10^{-23} \ J/K$,mass of oxygen molecule $= 5.31 \times 10^{-26} \ kg$).

If an electron has an energy such that its de-Broglie wavelength is $5500 \ \text{Å}$,then the energy value of that electron is $(h = 6.6 \times 10^{-34} \ \text{Js}, m_e = 9.1 \times 10^{-31} \ \text{kg})$.