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

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

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of 3 km/s will be
(1) 1 Å                     

(2) 0.66 Å

(3) 6.6 Å                 

(4) 66 Å

When the momentum of a proton is changed by an amount ${\mathrm{P}}_0$, the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was 
(a) ${\mathrm{P}}_0$                   (b) 100  ${\mathrm{P}}_0$
(c) 400 ${\mathrm{P}}_0$             (d) 4 ${\mathrm{P}}_0$

The de-Broglie wavelength of a neutron at 27 $°\mathrm{C}$ is $\mathrm{\lambda }$. What will be its wavelength at 927 $°\mathrm{C}$
(a) $\mathrm{\lambda }$ / 2                        (b) $\mathrm{\lambda }$ / 3
(c) $\mathrm{\lambda }$ / 4                        (d) $\mathrm{\lambda }$ / 9

An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

(1) Zero

(2) Infinity

(3) Equal to the kinetic energy of the proton

(4) Greater than the kinetic energy of the proton