For the moving ball of cricket, the correct statement about de-Broglie wavelength is:
1. It is not applicable for such big particle
2. \(\frac{h}{\sqrt{2mE}}\)
3. \(\sqrt{\frac{h}{2mE}}\)
4. \(\frac{h}{2mE}\)

The kinetic energy of an electron with de-Broglie wavelength of 0.3 nanometer is 
(1) 0.168 eV           

(2) 16.8 eV

(3) 1.68 eV             

(4) 2.5 eV

The wavelength of de-Broglie wave is 2$\mathrm{\mu }$m, then its momentum is (h = $6.63\times {10}^{-34}$ J-s) 
(a) $3.315\times {10}^{-28}$ kg-m/s            (b) $1.66\times {10}^{-28}$ kg-m/s
(c) $4.97\times {10}^{-28}$ kg-m/s              (d) $9.9\times {10}^{-28}$ kg-m/s

If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor 
(1) $\frac{1}{\sqrt{2}}$             

(2) $\sqrt{2}$

(3) $\frac{1}{2}$                

(4) 2

The energy that should be added to an electron to reduce its de Broglie wavelength from one nm to 0.5 nm is

(1) Four times the initial energy

(2) Equal to the initial energy

(3) Twice the initial energy

(4) Thrice the initial energy

The wavelength of the matter wave is independent of

(1) Mass                 

(2) Velocity

(3) Momentum       

(4) Charge

The energy of a photon of wavelength $\mathrm{\lambda }$ is given by
1. h$\mathrm{\lambda }$                      

2. ch$\mathrm{\lambda }$

3. $\mathrm{\lambda }$/hc                   

4.  hc/$\mathrm{\lambda }$

The rest mass of the photon is

(1) 0

(2)$\infty$

(3) Between 0 and $\infty$

(4) Equal to that of an electron

The momentum of a photon of energy hv will be

(1) hv     

(2) hv/c

(3) hvc     

(4) h/v

If the momentum of a photon is p, then its frequency is

(1) $\frac{\mathrm{ph}}{\mathrm{c}}$           

(2) $\frac{\mathrm{pc}}{\mathrm{h}}$

(3) $\frac{\mathrm{mh}}{\mathrm{c}}$           

(4) $\frac{\mathrm{mc}}{\mathrm{h}}$

where m is the rest mass of the photon