The de-Broglie wavelength associated with the particle of mass m moving with velocity v is

(1) h/mv                 

(2) mv/h 

(3) mh/v                 

(4) m/hv

A photon, an electron, and a uranium nucleus all have the same wavelength. The one with the most energy: 

(1) Is the photon

(2) Is the electron

(3) Is the uranium nucleus

(4) Depends upon the wavelength and the properties of the particle

When the kinetic energy of an electron is increased, the wavelength of the associated wave will
(1) Increase

(2) Decrease

(3) Wavelength does not depend on the kinetic energy

(4) None of the above

If the de-Broglie wavelengths for a proton and an alpha-particle are equal, then the ratio of their velocities will be:
1. \(4:1\)
2. \(2:1\)
3. \(1:2\)
4. \(1:4\)

The de-Broglie wavelength $\mathrm{\lambda }$ associated with an electron having kinetic energy E is given by the expression

(1) $\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$                           

(2) $\frac{2\mathrm{h}}{\mathrm{mE}}$

(3) 2mhE                               

(4) $\frac{2\sqrt{2\mathrm{mE}}}{\mathrm{h}}$

Dual nature of radiation is shown by:

(1) Diffraction and reflection

(2) Refraction and diffraction

(3) Photoelectric effect alone

(4) Photoelectric effect and diffraction

For the Bohr's first orbit of circumference $2\mathrm{\pi r}$, the de-Broglie wavelength of revolving electron will be

(a) $2\mathrm{\pi r}$                       (b) $\mathrm{\pi r}$
(c) $\frac{1}{2\mathrm{\pi r}}$                      (d) $\frac{1}{4\mathrm{\pi r}}$

An electron of mass m when accelerated through a potential difference V has de-Broglie wavelength $\mathrm{\lambda }$. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be

(1) $\mathrm{\lambda }\frac{\mathrm{m}}{\mathrm{M}}$                           

(2) $\mathrm{\lambda }\sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$

(3) $\mathrm{\lambda }\frac{\mathrm{M}}{\mathrm{m}}$                           

(4) $\mathrm{\lambda }\sqrt{\frac{M}{m}}$

What is the de-Broglie wavelength of the $\mathrm{\alpha }$-particle accelerated through a potential difference V 
(1) $\frac{0.287}{\sqrt{\mathrm{V}}}$ Å                 

(2) $\frac{12.27}{\sqrt{\mathrm{V}}}$ Å

(3) $\frac{0.101}{\sqrt{\mathrm{V}}}$ Å                 

(4) $\frac{0.202}{\sqrt{\mathrm{V}}}$ Å