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 P0, the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was
(a) P0 (b) 100 P0
(c) 400 P0 (d) 4 P0
The de-Broglie wavelength of a neutron at 27 °C is λ. What will be its wavelength at 927 °C
(a) λ / 2 (b) λ / 3
(c) λ / 4 (d) λ / 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. h√2mE
3. √h2mE
4. h2mE
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 2μm, then its momentum is (h = 6.63×10-34 J-s)
(a) 3.315×10-28 kg-m/s (b) 1.66×10-28 kg-m/s
(c) 4.97×10-28 kg-m/s (d) 9.9×10-28 kg-m/s
If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor
(1) 1√2
(2) √2
(3) 12
(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