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

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