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

Particle nature and wave nature of electromagnetic waves and electrons can be shown by 

(1) Electron has small mass, deflected by the metal sheet

(2) X-ray is diffracted, reflected by thick metal sheet

(3) Light is refracted and defracted

(4) Photoelectricity and electron microscopy

The speed of an electron having a wavelength of ${10}^{-10}$m is

(a) $7.25\times {10}^6$ m/s                   (b) $6.26\times {10}^6$ m/s 
(c) $5.25\times {10}^6$ m/s                   (d) $4.24\times {10}^6$ m/s

The kinetic energy of electron and proton is ${10}^{-32}$ J. Then the relation between their de-Broglie wavelengths is

(1) ${\mathrm{\lambda }}_{\mathrm{p}}<{\mathrm{\lambda }}_{\mathrm{e}}$                 

(2) ${\mathrm{\lambda }}_{\mathrm{p}}>{\mathrm{\lambda }}_{\mathrm{e}}$

(3) ${\mathrm{\lambda }}_{\mathrm{p}}={\mathrm{\lambda }}_{\mathrm{e}}$                 

(4) ${\mathrm{\lambda }}_{\mathrm{p}}=2{\mathrm{\lambda }}_{\mathrm{e}}$