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Physics MCQs for NEET — Practice Questions with Answers

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According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is $10^{- 9}$ m. The principle quantum number for this electron is $(r = 5.13 \times 10^{- 11} m)$

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Explanation

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The speed of an electron having a wavelength of $10^{- 10}$ m is

$7.25 \times 10^{6}$ m/s

$6.26 \times 10^{6}$ m/s

$5.25 \times 10^{6}$ m/s

$4.24 \times 10^{6}$ m/s

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Explanation

(a) By using

$λ_{electron} = \frac{h}{m_{e} v} \Rightarrow v = \frac{h}{m_{e} λ_{e}} = \frac{6.6 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 10^{- 10}} = 7.25 \times 10^{6} m / s$

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The kinetic energy of electron and proton is $10^{- 32}$ J. Then the relation between their de-Broglie wavelengths is

$λ_{p} < λ_{e}$

$λ_{p} > λ_{e}$

$λ_{p} = λ_{e}$

$λ_{p} = 2 λ_{e}$

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Explanation

(a) By using $λ = \frac{h}{\sqrt{2 mE}}$ E = $10^{- 32}$ J = Constant for both particles. Hence $λ \propto \frac{1}{\sqrt{m}}$ Since $m_{p} > m_{e}$ so $λ_{p} < λ_{e}$

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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

0.25 Å

0.5 Å

1.5 Å

2 Å

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Explanation

(b) By using $λ \propto \frac{1}{\sqrt{V}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{V_{2}}{V_{1}}} \Rightarrow \frac{10^{- 10}}{λ_{2}} = \sqrt{\frac{600}{150}} = 2 \Rightarrow λ_{2} = 0.5$ Å

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The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of 3 km/s will be

1 Å

0.66 Å

6.6 Å

66 Å

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Explanation

(b) $λ = \frac{h}{{mv}_{rms}} \Rightarrow λ = \frac{6.6 \times 10^{- 34}}{2 \times 1.67 \times 10^{- 27} \times 3 \times 10^{3}} = 0.66$ Å

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When the momentum of a proton is changed by an amount $P_{0}$ , the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was

$P_{0}$

100 $P_{0}$

400 $P_{0}$

4 $P_{0}$

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Explanation

(c) $λ \propto \frac{1}{p} \Rightarrow \frac{∆ p}{p} = - \frac{∆ λ}{λ} \Rightarrow |\frac{∆ p}{p}| = |\frac{∆ λ}{λ}| \Rightarrow \frac{p_{0}}{p} = \frac{0.25}{100} = \frac{1}{400} \Rightarrow p = 400 p_{0}$

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The de-Broglie wavelength of a neutron at 27 $° C$ is $λ$ . What will be its wavelength at 927 $° C$

$λ$ / 2

$λ$ / 3

$λ$ / 4

$λ$ / 9

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Explanation

(a) $λ_{neutron} \propto \frac{1}{\sqrt{T}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{T_{2}}{T_{1}}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{(273 + 927)}{(273 + 27)}} = \sqrt{\frac{1200}{300}} = 2 \Rightarrow λ_{2} = \frac{λ}{2}$

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An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

Zero

Infinity

Equal to the kinetic energy of the proton

Greater than the kinetic energy of the proton

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Explanation

(d) $λ = \frac{h}{\sqrt{2 mE}} \Rightarrow E \propto \frac{1}{\sqrt{m}} (λ = constant) ∵ m_{e} < m_{p} so E_{e} > E_{p}$

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For moving ball of cricket, the correct statement about de-Broglie wavelength is

It is not applicable for such big particle

$\frac{h}{\sqrt{2 mE}}$

$\sqrt{\frac{h}{2 mE}}$

$\frac{h}{2 mE}$

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Explanation

(b) For moving ball with velocity v-

$λ = \frac{h}{p} = \frac{λ}{\sqrt{2 m E}}$

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The kinetic energy of an electron with de-Broglie wavelength of 0.3 nanometer is

0.168 eV

16.8 eV

1.68 eV

2.5 eV

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Explanation

(b) $λ = \frac{h}{\sqrt{2 mE}} \Rightarrow E = \frac{h^{2}}{2 {mλ}^{2}} = \frac{{(6.6 \times 10^{- 34})}^{2}}{2 \times 9.1 \times 10^{- 31} \times {(0.3 \times 10^{- 9})}^{2}} = 2.65 \times 10^{- 18} J = 16.8 eV$

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