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Physics NEET Questions — Free MCQ Practice

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Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the accelerating voltage is reduced to 6.25 kV, the resolving power of the microscope would:

A.

Increase to 4 time

B.

Increases to 2 times

C.

Decrease to half

D.

Decrease to 1/4th

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Explanation

Resolving power ( R.P. ) $\propto \frac{1}{λ}$

By de-Broglie relation-

$λ = \frac{h}{\sqrt{2 m e V}}$

So. R.P. $\propto \sqrt{V}$

So new R.P. = $\frac{1}{2} t h o f i n i t i a l R . P .$

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When the light of wavelength $λ$ is used in a photoelectric experiment, the maximum kinetic energy of emitted electrons is found to be K. What will the maximum kinetic energy of the photoelectrons when the light of wavelength $\frac{λ}{2}$ is used instead of $λ$ :

A.

Zero

B.

Non zero but less than 2 K

C.

2K

D.

More than 2K

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Explanation

$When wavelength λ is used : \frac{hc}{λ} = K + ϕ ϕ = \frac{hc}{λ} - K When wavelength \frac{λ}{2} is used : \frac{2 hc}{λ} = K' + ϕ = K' + \frac{hc}{λ} - K K' = \frac{2 hc}{λ} - \frac{hc}{λ} + K = \frac{hc}{λ} + K = K + ϕ + K = 2 K + ϕ K' is more than 2 K .$

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The concept of matter-wave was given by:

A.

de-Broglie

B.

Einstein

C.

Maxwell

D.

Bohr

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Explanation

Each moving particle consists of a wave. This wave is called matter-wave of De-Broglie wave.

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When the light of wavelength $λ = 3100 \overset{o}{A}$ is made to fall on metal in a photoelectric experiment, the maximum kinetic energy of emitted electrons is found to be 2.5eV. The work function of the metal is:

A.

2.5eV

B.

3.1eV

C.

0.9eV

D.

1.5eV

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Explanation

$h ν = K E + ϕ ϕ = \frac{h c}{λ} - K E = \frac{12400}{3100} - 2.5 = 1.5 e V$

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If the momentum of a photon is p, then its energy is :

A.

pc

B.

$\frac{1}{2} pc$

C.

$\frac{p^{2}}{2 c}$

D.

$\frac{1}{4} pc$

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Explanation

$(1) Energy of photon = \frac{hc}{λ} = p x c$

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For non-relativistic speeds, the wavelength associated with an electron and its kinetic energy E are related as:

A.

$λ \propto E$

B.

$λ \propto E^{- 1}$

C.

$λ \propto E^{1 / 2}$

D.

$λ \propto E^{- 1 / 2}$

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Explanation

$(4) λ = \frac{h}{\sqrt{2 m (K . E .)}} \Rightarrow λ \propto \frac{1}{\sqrt{E}}$

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An electron at rest is acceleration by a potential difference of 600V. The de-Broglie wavelength associated with the electron is:

A.

$1 \overset{o}{A}$

B.

$2 \overset{o}{A}$

C.

$3 \overset{o}{A}$

D.

$0.5 \overset{o}{A}$

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Explanation

$(3) λ = \sqrt{\frac{150}{V}} \overset{o}{A} (for electron) λ = \sqrt{\frac{150}{600}} = \frac{1}{2} \overset{o}{A}$

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The de-Broglie wavelength of a body of mass 1 kg moving with a velocity of 2000 m/s is

A.

$3.32 x 10^{- 27} \overset{o}{A}$

B.

$1.5 x 10^{7} \overset{o}{A}$

C.

$0.55 x 10^{- 22} \overset{o}{A}$

D.

None of these

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Explanation

$(1) λ = \frac{h}{mv} = \frac{6.6 x 10^{- 34}}{1 x 2 x 10^{3}} = 3.3 x 10^{- 37} = 3.3 x 10^{- 27} \overset{o}{A}$

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A particle of mass M at rest decays into two particles of

masses $m_{1} a n d m_{2}$ having non-zero velocities.

The ratio of de-Broglie wavelength is:

A.

$\frac{m_{1}}{m_{2}}$

B.

$\frac{m_{2}}{m_{1}}$

C.

1

D.

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

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Explanation

$(3) 0 = \vec{p_{1}} + \vec{p_{2}} |\vec{p_{2}}| = |\vec{p_{1}}| λ = \frac{h}{p} λ_{1} = λ_{2}$

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The de-Brogile wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is

A.

$\frac{h}{\sqrt{m k T}}$

B.

$\frac{h}{\sqrt{3 m k T}}$

C.

$\frac{2 h}{\sqrt{3 m k T}}$

D.

$\frac{2 h}{\sqrt{m k T}}$

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Explanation

(b) Thinking Processs de-Brogile wavelength associated with a moving particle can be given as

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m (K . E)}}$

At thermal equilibrium,temperature of neutron and heavy water will be same. This common temperature is given as, T. Also, we know that, kinetic energy of a particle

$K E = \frac{p^{2}}{2 m}$

where, p=momentum of the particle

m=mass of the particle

Kinetic energy of the neutron is

$K E = \frac{3}{2} k T$

$∴$ de-Brogile wavelength of the neutron

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m (K . E)}}$

$= \frac{h}{\sqrt{2 m \times \frac{3}{2} k T}} = \frac{h}{\sqrt{3 m k T}}$

Physics MCQs for NEET — Practice Questions with Answers

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