# The de-Broglie wavelength $\mathrm{\lambda }$ associated with an electron having kinetic energy E is given by the expression (1) $\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$                            (2) $\frac{2\mathrm{h}}{\mathrm{mE}}$ (3) 2mhE                                (4) $\frac{2\sqrt{2\mathrm{mE}}}{\mathrm{h}}$

Subtopic:  De-broglie Wavelength |
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Dual nature of radiation is shown by:

(1) Diffraction and reflection

(2) Refraction and diffraction

(3) Photoelectric effect alone

(4) Photoelectric effect and diffraction

Subtopic:  Photoelectric Effect: Experiment |
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An electron of mass m when accelerated through a potential difference V has de-Broglie wavelength $\mathrm{\lambda }$. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be

(1) $\mathrm{\lambda }\frac{\mathrm{m}}{\mathrm{M}}$

(2) $\mathrm{\lambda }\sqrt{\frac{\mathrm{m}}{\mathrm{M}}}$

(3) $\mathrm{\lambda }\frac{\mathrm{M}}{\mathrm{m}}$

(4) $\mathrm{\lambda }\sqrt{\frac{M}{m}}$

Subtopic:  De-broglie Wavelength |
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What is the de-Broglie wavelength of the $\mathrm{\alpha }$-particle accelerated through a potential difference V
(1) $\frac{0.287}{\sqrt{\mathrm{V}}}$ Å

(2) $\frac{12.27}{\sqrt{\mathrm{V}}}$ Å

(3) $\frac{0.101}{\sqrt{\mathrm{V}}}$ Å

(4) $\frac{0.202}{\sqrt{\mathrm{V}}}$ Å

Subtopic:  De-broglie Wavelength |
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How much energy should be added to an electron to reduce its de-Broglie wavelength from $$10^{-10}$$ m to $$0.5\times10^{-10}$$${}^{}$ m?
1. Four times the initial energy.
2. Thrice the initial energy.
3. Equal to the initial energy.
4. Twice the initial energy.

Subtopic:  De-broglie Wavelength |
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The de-Broglie wavelength of an electron having 80eV of energy is nearly
(1eV =$1.6×{10}^{-19}$ J, Mass of electron = $9×{10}^{-31}$Kg Plank’s constant = $6.6×{10}^{-34}$ J-sec)

(a) 140 Å                      (b) 0.14 Å
(c) 14 Å                        (d) 1.4 Å

Subtopic:  De-broglie Wavelength |
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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

Subtopic:  De-broglie Wavelength |
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If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
(1) Energy

(2) Momentum

(3) Velocity

(4) Angular momentum

Subtopic:  De-broglie Wavelength |
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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) $\mathrm{\lambda }\propto \mathrm{p}$

Subtopic:  De-broglie Wavelength |
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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

Subtopic:  Particle Nature of Light |
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