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

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 de-Broglie wavelength of a particle moving with a velocity $2.25\times {10}^8$ m/s is equal to the wavelength of the photon. What is the ratio of the kinetic energy of the particle to the energy of the photon? (velocity of light is $3\times {10}^8$ m/s)

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

The de-Broglie wavelength of a particle accelerated with 150 volt potential is ${10}^{-10}$ m. If it is accelerated by 600 volts p.d., its wavelength will be
(1) 0.25 Å                 

(2) 0.5 Å

(3) 1.5 Å                   

(4) 2 Å