The circumference of the Bohr orbit for the H atom is related to the de Broglie wavelength associated with the electron revolving around the orbit by the following relation:

$1.$ $2\mathrm{\pi r}=\mathrm{n\lambda }$

$2.$ $\mathrm{\pi r}$ $=$ $2\mathrm{n\lambda }$

$3.$ $\mathrm{mvr}$ $=$ $\mathrm{n\lambda }$

$4.$ $\mathrm{vr}$ $=$ $2\mathrm{n\lambda }$

Subtopic:  Bohr's Theory | De Broglie Equation |
80%
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The largest de Broglie wavelength among the following (all have equal velocity) is:

1.  ${\mathrm{CO}}_{2}$ molecule

2.  ${\mathrm{NH}}_{3}$ molecule

3.  Electron

4.  Proton

Subtopic:  De Broglie Equation |
78%
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The wavelength of an electron moving with a velocity of 2.05 × 107 m s-1  would be:

$$1 .$$ $$4 . 65$$ $$\times$$ $$\left(10\right)^{- 12}$$ $$m$$

$$2 .$$ $$3 . 55$$ $$\times$$ $$\left(10\right)^{-11}$$ $$m$$

$$3 .$$ $$2 . 34$$ $$\times$$ $$\left(10\right)^{11}$$ $$m$$

$$4 .$$ $$6 . 43$$ $$\times$$ $$\left(10\right)^{ -11}$$ $$m$$

Subtopic:  De Broglie Equation |
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If the velocity of the electron is 1.6 × 106 ${\mathrm{ms}}^{-1}$. The de Broglie wavelength associated with this electron is:

$1.$ $590$ $\mathrm{pm}$
$2.$ $455$ $\mathrm{pm}$
$3.$ $512$ $\mathrm{pm}$
$4.$ $401$ $\mathrm{pm}$

Subtopic:  De Broglie Equation |
72%
From NCERT
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The graphs that represents the variation of momentum of particle with de-Broglie wavelength is:

 1 2 3 4

Subtopic:  De Broglie Equation |
67%
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Incorrect statement among the following is:

 1 The uncertainty principle is$$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$$ 2 Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry, and a more balanced arrangement. 3 The energy of the 2s orbital is less than the energy of the 2p orbital in the case of hydrogen-like atoms. 4 De-Broglie's wavelength is given by$$\lambda=\frac{h}{m v}$$, where m= mass of the particle, v = group velocity of the particle.

Subtopic:  De Broglie Equation |
67%
From NCERT
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The velocity associated with a proton moving at a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, then the wavelength associated with this velocity would be:

$1.$ $1.52$ $×{10}^{-38}$ $\mathrm{m}$
$2.$ $2.54$ $×$ ${10}^{-32}$ $\mathrm{m}$
$3.$ $1.52$ $×$ ${10}^{-36}$ $\mathrm{m}$
$4.$ $3.19$ $×$ ${10}^{-34}$ $\mathrm{m}$

Subtopic:  De Broglie Equation |
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From NCERT
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The kinetic energy of an electron is $$3.0 \times 10^{-25}~ \mathrm J.$$ Its wave length would be:

 1 $$8.96 \times 10^{-7}~ \mathrm m$$ 2 $$4.37 \times 10^{-6}~ \mathrm m$$ 3 $$1.32 \times 10^{-7}~ \mathrm m$$ 4 $$2.89 \times 10^{-4}~ \mathrm m$$
Subtopic:  De Broglie Equation |
55%
From NCERT
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The characteristic velocity associated with a neutron that has a wavelength of 800 pm is :

 1 496 m s-1 2 567 m s-1 3 494 cm s-1 4 501 m s-1
Subtopic:  De Broglie Equation |
51%
From NCERT
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