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

A particle which has zero rest mass and non-zero energy and momentum must travel with a speed:

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

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

How much energy should be added to an electron to reduce its de-Broglie wavelength from ${10}^{-10}$ m to $0.5\times {10}^{-10}$m?
1. Four times the initial energy.
2. Thrice the initial energy.
3. Equal to the initial energy.
4. Twice the initial energy.

The de-Broglie wavelength of an electron having 80eV of energy is nearly
(1eV =$1.6\times {10}^{-19}$ J, Mass of electron = $9\times {10}^{-31}$Kg Plank’s constant = $6.6\times {10}^{-34}$ J-sec)

(a) 140 Å                      (b) 0.14 Å
(c) 14 Å                        (d) 1.4 Å