Dual Nature of Radiation and Matter — Page 3 | JEE Physics

Q21

This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement 1 : Davisson - Germer experiment established the wave nature of electrons. Statement 2 : If electrons have wave nature, they can interfere and show diffraction.

A Statement 1 is true, Statement 2 is false

B Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1.

C Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1.

D Statement 1 is false, Statement 2 is true.

Correct Answer

Option B

Solution

Davisson-Germer experiment showed that electron beams can undergo diffraction when passed through atomic crystals.

This shows the wave nature of electrons as waves can exhibit interference and diffraction.

Q22

Radiation of wavelength

\lambda ,

is incident on a photocell. The fastest emitted electron has speed

v.

If the wavelength is changed to

{{3\lambda } \over 4},

the speed of the fastest emitted electron will be:

A

= v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}

B

= v{\left( {{3 \over 4}} \right)^{{1 \over 2}}}

C

> v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}

D

< v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}

Correct Answer

Option C

Solution

h{v_0}^2 - h{v_0} = {1 \over 2}m{v^2}

$\therefore$

{4 \over 3}h{v_0} - h{v_0} = {1 \over 2}mv{'^2}

$\therefore$

{{v{'^2}} \over {{v^2}}} = {{{4 \over 3}v - {v_0}} \over {v - {v_0}}}

$\therefore$

v' = v\sqrt {{{{4 \over 3}v - {v_0}} \over {v - {v_0}}}}

$\therefore$

v' > v\sqrt {{4 \over 3}}

Q23

A particle moving with kinetic energy E has de Broglie wavelength

\lambda

. If energy

\Delta

E is added to its energy, the wavelength become

\lambda

/2. Value of

\Delta

E, is :

A E

B 3E

C 2E

D 4E

Correct Answer

Option B

Solution

\lambda = {h \over {\sqrt {2mE} }}

Also,

{h \over {\sqrt {2m\left( {E + \Delta E} \right)} }}

=

{\lambda \over 2}

$\therefore$

{{E + \Delta E} \over E} = 4

$\Rightarrow$

\Delta

E = 3E

Q24

When radiation of wavelength

\lambda

is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is :

A 2

\lambda

B 4

\lambda

C 8

\lambda

D 6

\lambda

Correct Answer

Option B

Solution

{V_s} = hv - \phi

4.8 = {{hc} \over \lambda } - \phi

..... (i)

1.6 = {{hc} \over {2\lambda }} - \phi

..... (ii) Using above equation (i) - (ii)

3.2 = {{hc} \over \lambda } - {{hc} \over {2\lambda }}

3.2 = {{hc} \over {2\lambda }}

..... (iii)

\left[ {\lambda = {{hc} \over {6.4}}} \right]

Put in equation (ii) $\phi$ = 1.6

{{hc} \over {{\lambda _{th}}}} = 1.6

{\lambda _{th}} = {{hc} \over {1.6}}

= \left( {{{hc} \over {6.4}}} \right) \times 4 = 4\lambda

Q25

A proton, a neutron, an electron and an

\alpha

particle have same energy. If

\lambda

p,

\lambda

n,

\lambda

e and

\lambda

a are the de Broglie's wavelengths of proton, neutron, electron and

\alpha

particle respectively, then choose the correct relation from the following :

A

\lambda

p =

\lambda

n >

\lambda

e >

\lambda

a

B

\lambda

a n p e

C

\lambda

e p =

\lambda

n >

\lambda

a

D

\lambda

e =

\lambda

p =

\lambda

n =

\lambda

a

Correct Answer

Option B

Solution

\lambda=\frac{h}{p}=\frac{h}{\sqrt{(2 m K E)}}

$\lambda \propto \dfrac{1}{\sqrt{m}}$ as all the particles have same KE.

Since $m_e \lambda_p > \lambda_{\mathrm{n}} > \lambda_\alpha$ $

Q26

With reference to the observations in photo-electric effect, identify the correct statements from below : (A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. (B) The value of saturation current increases on moving the source of light away from the metal surface. (C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. (D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. (E) Existence of threshold wavelength can not be explained by wave nature of light/ electromagnetic waves. Choose the correct answer from the options given below :

A (A) and (B) only

B (A) and (E) only

C (C) and (E) only

D (D) and (E) only

Correct Answer

Option B

Solution

$\because$

{1 \over 2}mv_m^2 = h\nu - \phi

\Rightarrow v_m^2

varies linearly with frequency. And, threshold wavelength can be explained by particle nature of light.

Q27

Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of

\lambda_0

. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :

A 3

\lambda_0

B 9

\lambda_0

C

\dfrac{\lambda_0}{\sqrt2}

D

\dfrac{\lambda_0}{2}

Correct Answer

Option C

Solution

When electron is accelerated through potential difference $V$ , then

\begin{aligned} & \text{ K.E. }=\mathrm{eV} \\\\ & \Rightarrow \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\\\ & \therefore \lambda \alpha \frac{1}{\sqrt{\mathrm{V}}} \\\\ & \therefore \frac{\lambda}{\lambda_0}=\sqrt{\frac{20}{40}} \\\\ & \therefore \lambda=\frac{\lambda_0}{\sqrt{2}} \end{aligned}

Q28

A sub-atomic particle of mass

10^{-30} \mathrm{~kg}

is moving with a velocity

2.21 \times 10^6 \mathrm{~m} / \mathrm{s}

. Under the matter wave consideration, the particle will behave closely like

\qquad

\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)

A X-rays

B Infra-red radiation

C Gamma rays

D Visible radiation

Correct Answer

Option A

Solution

To find the matter wave behavior of the particle, we use the de Broglie wavelength formula:

\lambda = \frac{h}{mv}

where:

h = 6.63 \times 10^{-34} \, \mathrm{J \cdot s}

is Planck’s constant,

m = 10^{-30} \, \mathrm{kg}

is the mass of the particle, and

v = 2.21 \times 10^6 \, \mathrm{m/s}

is its velocity. Follow these steps: Substitute the values into the formula:

\lambda = \frac{6.63 \times 10^{-34}}{(10^{-30})(2.21 \times 10^6)}

Calculate the denominator:

10^{-30} \times 2.21 \times 10^6 = 2.21 \times 10^{-24}

Now compute the wavelength:

\lambda \approx \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}} \approx 3 \times 10^{-10} \, \mathrm{m}

This wavelength, roughly

3 \times 10^{-10} \, \mathrm{m}

(or 0.3 nm), lies within the X-ray portion of the electromagnetic spectrum, since X-rays typically have wavelengths in the range of about 0.01 nm to 10 nm.

Therefore, under the matter wave consideration, the particle will behave closely like: Option A: X-rays.

Q29

The speed of electrons in a scanning electron microscope is 1

\times

107 ms-1. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of :

A

{1 \over {1837}}

B 1837

C

{\sqrt {1837} }

D

{1 \over {\sqrt {1837} }}

Correct Answer

Option B

Solution

Resolving power (RP) $\propto$

{1 \over \lambda }

We know, de-Broglie wavelength

\lambda = {h \over {mv}}

$\therefore$ RP $\propto$

{mv \over {h}}

$\therefore$

{{R{P_e}} \over {R{P_p}}} = {{{m_e}} \over {{m_p}}} = 1837

Q30

When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to :

A 1.02 eV

B 0.81 eV

C 0.61 eV

D 0.52 eV

Correct Answer

Option C

Solution

K1 =

{{hc} \over {500}} - {\phi _0}

K2 =

{{hc} \over {200}} - {\phi _0}

$\because$ K2 = 3K1 $\Rightarrow$

3\left[ {{{hc} \over {500}} - {\phi _0}} \right] = \left[ {{{hc} \over {200}} - {\phi _0}} \right]

$\Rightarrow$ $\phi$ 0 = 0.61 eV