A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of $3\times {10}^6$ $m{s}^{-1}$. What will be the velocity of the particle? (mass of electrons = 9 . 1 × 10^{- 31} kg )

1.	\(2.7 \times 10^{-18} \mathrm{~ms}^{-1}\)
2.	\(9 \times 10^{-2} \mathrm{~ms}^{-1}\)
3.	\(3 \times 10^{-31} \mathrm{~ms}^{-1}\)
4.	\(2.7 \times 10^{-21} \mathrm{~ms}^{-1}\)

Waves are associated with matter only:

A particle which has zero rest mass and non-zero energy and momentum must travel with a speed:

If the de-Broglie wavelengths for a proton and an alpha-particle are equal, then the ratio of their velocities will be:
1. 4 : 1
2. 2 : 1
3. 1 : 2
4. 1 : 4

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.

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

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)

A proton and an $\mathrm{\alpha }$-particle are accelerated through a potential difference of 100 V. What is the ratio of the wavelength associated with the proton to that of the alpha-particle?
1. $\sqrt{2}:1$              

2. 2:1

3. $2\sqrt{2}:1$             

4. $\frac{1}{2\sqrt{2}}:1$

What is the momentum of a photon in an X-ray beam of 10^-10 meter wavelength?

The energy of a quanta of frequency ${10}^{15}$ Hz and $\mathrm{h}=6.6\times {10}^{-34}\mathrm{ J}-\sec$ will be:

1. $6.6\times {10}^{-19}$ $\mathrm{J}$

$\mathrm{}$2.$6.6\times {10}^{-12}$ $\mathrm{J}$

3. $6.6\times {10}^{-49}$ $\mathrm{J}$

$\mathrm{}$4.$6.6\times {10}^{-41}$ $\mathrm{J}$