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Physics MCQs for NEET — Practice Questions with Answers

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A proton and an $α$ -particle are accelerated through a potential difference of 100 V. The ratio of the wavelength associated with the proton to that associated with an $α$ -particle is

$\sqrt{2} : 1$

2:1

$2 \sqrt{2} : 1$

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

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Explanation

(c) $λ = \frac{h}{\sqrt{2 mQV}} \Rightarrow λ \propto \frac{1}{\sqrt{mQ}} \Rightarrow \frac{λ_{p}}{λ_{α}} = \sqrt{\frac{m_{α} Q_{α}}{m_{p} Q_{p}}} = \sqrt{\frac{4 m_{p} \times 2 Q_{p}}{m_{p} \times Q_{p}}} = 2 \sqrt{2}$

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The wavelength of de-Broglie wave is 2 $μ$ m, then its momentum is (h = $6.63 \times 10^{- 34}$ J-s)

$3.315 \times 10^{- 28}$ kg-m/s

$1.66 \times 10^{- 28}$ kg-m/s

$4.97 \times 10^{- 28}$ kg-m/s

$9.9 \times 10^{- 28}$ kg-m/s

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Explanation

(a) $λ = \frac{h}{p} \Rightarrow p = \frac{h}{λ} = \frac{6.63 \times 10^{- 34}}{2 \times 10^{- 6}} = 3.31 \times 10^{- 28} kg m / \sec$

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Davission and Germer experiment proved

Wave nature of light

Particle nature of light

Both (a) and (b)

Neither (a) nor (b)

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Explanation

(d) Davission and Germer proved the wave nature of electron by performing an experiment.

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If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

$\frac{1}{2}$

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Explanation

(a) $λ = \frac{h}{\sqrt{2 mE}} \Rightarrow λ \propto \frac{1}{\sqrt{E}}$

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The energy that should be added to an electron to reduce its de Broglie wavelength from one nm to 0.5 nm is

Four times the initial energy

Equal to the initial energy

Twice the initial energy

Thrice the initial energy

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Explanation

(d) $λ = \frac{h}{\sqrt{2 mE}}; \frac{λ'}{λ} = \sqrt{\frac{E}{E'}} \Rightarrow \frac{E}{E} = {(\frac{0.5}{1})}^{2} \Rightarrow E' = \frac{E}{0.25} = 4 E$

The energy should be added to decrease wavelength.
= E' - E = 3E

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The wavelength of the matter wave is independent of

Mass

Velocity

Momentum

Charge

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Explanation

(d) $λ = \frac{h}{p}$

So, wavelength is independent of charge.

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The energy of a photon of wavelength $λ$ is given by

h $λ$

ch $λ$

$λ$ /hc

hc/ $λ$

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Explanation

(d) $E = h v = \frac{h e}{λ}$

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The rest mass of the photon is

$\infty$

Between 0 and $\infty$

Equal to that of an electron

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Explanation

The rest mass of photon is zero.

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The momentum of the photon of wavelength 5000Å will be

$1.3 \times 10^{- 27}$ kg-m/sec

$1.3 \times 10^{- 28}$ kg-m/sec

$4 \times 10^{29}$ kg-m/sec

$4 \times 10^{- 18}$ kg-m/sec

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Explanation

(a) $p = \frac{h}{λ} = \frac{6.6 \times 10^{- 34}}{(5000 \times 10^{- 10})} = 1.3 \times 10^{- 27} kg - m / s$

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The momentum of a photon of energy hv will be

hv/c

hvc

h/v

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Explanation

(b) $p = \frac{E}{c} = \frac{hv}{c}$

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