Questions in Dual Nature of Radiation and Matter

Select	Question
	An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is
	For moving ball of cricket, the correct statement about de-Broglie wavelength is
	Photon and electron are given same energy $({10^{ - 20}}J)$. Wavelength associated with photon and electron are ${\lambda _{Ph}}$ and ${\lambda _{el}}$ then correct statement will be
	The kinetic energy of an electron with de-Broglie wavelength of 0.3 nanometer is
	A proton and an ?-particle are accelerated through a potential difference of 100 V. The ratio of the wavelength associated with the proton to that associated with an $\alpha$ -particle is
	The wavelength of de-Broglie wave is $2\mu m$, then its momentum is ($h = 6.63 \times 10^{–34} J-s$)
	de-Broglie wavelength of a body of mass 1 kg moving with velocity of 2000 m/s is
	The kinetic energy of an electron is 5 eV. Calculate the de-Broglie wavelength associated with it ($h = 6.63 \times 10^{–34} J-s$, $m_e = 9.1 \times 10^{–31}$ kg)
	The wavelength associated with an electron accelerated through a potential difference of 100 V is nearly
	The de-Broglie wavelength $\lambda$