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 \(0.5\times10^{-10}\)${\mathrm{}}^{}$ 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

If the following particles are moving at the same velocity, then which among them will have the maximum de-Broglie wavelength?
1. Neutron               

2. Proton

3. $\mathrm{\beta }$-particle             

4. $\mathrm{\alpha }$-particle

If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same 
(1) Energy             

(2) Momentum

(3) Velocity             

(4) Angular momentum

The de-Broglie wavelength is proportional to 
(1) $\mathrm{\lambda }\propto \frac{1}{\mathrm{v}}$               

(2) $\mathrm{\lambda }\propto \frac{1}{m}$

(3) $\mathrm{\lambda }\propto \frac{1}{p}$               

(4) $\mathrm{\lambda }\propto \mathrm{p}$