Dual Nature of Radiation and Matter — Page 9 | JEE Physics

Q81

The de Broglie wavelengths of a proton and an

\alpha

particle are

\lambda

and

2 \lambda

respectively. The ratio of the velocities of proton and

\alpha

particle will be :

A

8: 1

B

1: 2

C

1: 8

D

4: 1

Correct Answer

Option A

Solution

To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength.

The de Broglie's wavelength formula is given by:

\lambda = \frac{h}{p}

where: $\lambda$ is the de Broglie wavelength, $h$ is the Planck constant, and $p$ is the momentum of the particle.

The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$ :

p = mv

So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:

\lambda = \frac{h}{mv}

For the proton (let's use subscript $p$ for proton), the wavelength is $\lambda$ :

\lambda_p = \frac{h}{m_p v_p}

For the $\alpha$ particle (let's use subscript $\alpha$ for alpha), the wavelength is $2\lambda$ :

2\lambda = \frac{h}{m_\alpha v_\alpha}

We are interested in finding the ratio of the velocities $\dfrac{v_p}{v_\alpha}$ . Using the given data about wavelengths:

\lambda_p = \lambda

\lambda_\alpha = 2\lambda

Using the de Broglie equation for both particles:

\frac{h}{m_p v_p} = \lambda

\frac{h}{m_\alpha v_\alpha} = 2\lambda

Dividing the second equation by the first equation gives us:

\frac{\frac{h}{m_\alpha v_\alpha}}{\frac{h}{m_p v_p}} = \frac{2\lambda}{\lambda}

\frac{m_p v_p}{m_\alpha v_\alpha} = \frac{2}{1}

\frac{v_p}{v_\alpha} = \frac{2m_\alpha}{m_p}

Since we know an $\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\alpha$ particle is roughly four times the mass of a proton ( $m_\alpha \approx 4m_p$ ).

Substituting $m_\alpha$ with $4m_p$ in the equation:

\frac{v_p}{v_\alpha} = \frac{2(4m_p)}{m_p}

\frac{v_p}{v_\alpha} = 2 \cdot 4

\frac{v_p}{v_\alpha} = 8

Therefore, the ratio of the velocities of proton to $\alpha$ particle is $8:1$ , which corresponds to Option A.

Q82

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

A

\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_\alpha

B

\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}

C

\lambda_{\mathrm{e}}>\lambda_\alpha>\lambda_{\mathrm{p}}

D

\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}<\lambda_\alpha

Correct Answer

Option B

Solution

To determine the relationship between the de-Broglie wavelengths of a proton, an electron, and an alpha particle with the same energies, we need to use the de-Broglie wavelength formula:

\lambda = \frac{h}{p}

where $h$ is Planck's constant and $p$ is the momentum of the particle.

For particles with the same kinetic energy $E$ , we have:

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

Solving for $p$ :

p = \sqrt{2mE}

Substituting this into the de-Broglie equation, we get:

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

Since all three particles have the same energy $E$ , the de-Broglie wavelength is inversely proportional to the square root of the mass $m$ :

\lambda \propto \frac{1}{\sqrt{m}}

The masses of the particles are as follows: Mass of proton $m_p$ is approximately $1.67 \times 10^{-27}$ kg Mass of electron $m_e$ is approximately $9.11 \times 10^{-31}$ kg Mass of alpha particle $m_{\alpha}$ is approximately $4 \times 1.67 \times 10^{-27}$ kg (since it has 2 protons and 2 neutrons) Comparatively: Mass of alpha particle $m_{\alpha}$ is the largest.

Mass of proton $m_p$ is intermediate.

Mass of electron $m_e$ is the smallest.

Therefore, the de-Broglie wavelength will be: Largest for the electron ( $\lambda_{\mathrm{e}}$ ) Intermediate for the proton ( $\lambda_{\mathrm{p}}$ ) Smallest for the alpha particle ( $\lambda_{\alpha}$ ) Hence, the correct order is: Option B: $$\lambda_\alpha

Q83

Sodium and copper have work functions

2.3

eV

and

4.5

eV

respectively. Then the ratio of the wavelengths is nearest to

A

1:2

B

4:1

C

2:1

D

1:4

Correct Answer

Option C

Solution

We know that work function is the energy required and energy

E = h\upsilon

$\therefore$

{{{E_{NA}}} \over {{E_{Cu}}}} = {{h{\upsilon _{Na}}} \over {h{\upsilon _{cu}}}} = {{{\lambda _{cu}}} \over {{\lambda _{Na}}}}

\left[ {\,\,} \right.

as

{\,\,\,\upsilon \propto {1 \over \lambda }}

for light

\left. {\,\,} \right]

$\therefore$

{{{\lambda _{Na}}} \over {{\lambda _{Cu}}}} = {{{E_{Cu}}} \over {{E_{Na}}}} = {{4.5} \over {2.3}} \approx {2 \over 1}

Q84

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

A 1 : 1

B 2 : 1

C 4 : 1

D 1 : 4

Correct Answer

Option B

Solution

3.8 = 0.6 + {1 \over 2}mv_1^2

1.4 = 0.6 + {1 \over 2}mv_2^2

\Rightarrow {{v_1^2} \over {v_2^2}} = {{3.2} \over {0.8}} = {4 \over 1}

\Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 1}

Q85

According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal

Vs

the frequency, of the incident radiation gives as straight the whose slope

A depends both on the intensity of the radiation and the metal used

B depends on the intensity of the radiation

C depends on the nature of the metal used

D is the same for the all metals and independent of the intensity of the radiation

Correct Answer

Option D

Solution

Einstein's photoelectric equation is given by

K_{\max} = h\nu - \phi,

where $K_{\max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident radiation, and $\phi$ is the work function of the metal (the minimum energy needed to eject an electron).

If you plot $K_{\max}$ against $\nu$ , you will get a straight line with slope $h$ (Planck's constant) and y-intercept $-\phi$ (the negative of the work function).

The slope of the line (which is $h$ ) does not depend on the intensity of the radiation or the type of metal used.

Instead, it is a universal constant.

Therefore, the correct answer is: Option D: The slope is the same for all metals and independent of the intensity of the radiation.

Q86

The threshold frequency for a metallic surface corresponds to an energy of

6.2

eV

and the stopping potential for a radiation incident on this surface is

5V.

The incident radiation lies in

A ultra-violet region

B infra-red region

C visible region

D

x

-ray region

Correct Answer

Option A

Solution

The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation

E = eV,

where e is the elementary charge. The elementary charge, e, is approximately equal to

1.602 \times 10^{-19}

C. So the energy of the incident radiation is

E = 1.602 \times 10^{-19} \, \text{C} \times 5 \, \text{V} = 8.01 \times 10^{-19} \, \text{J}.

We convert this to electron volts (eV) by using the conversion factor

1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}.

So

E = \frac{8.01 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 5 \, \text{eV}.

This is less than the threshold energy of

6.2 \, \text{eV},

which means that the incident radiation does not have enough energy to overcome the work function of the metal and thus cannot cause photoelectrons to be emitted.

The regions of the electromagnetic spectrum are generally classified by energy as follows: Infrared: Less than about

1.24 \, \text{eV}

Visible: Between about

1.24 \, \text{eV}

and

3.1 \, \text{eV}

Ultraviolet: Between about

3.1 \, \text{eV}

and

124 \, \text{eV}

X-rays: Greater than about

124 \, \text{eV}

Therefore, the incident radiation falls in the ultraviolet region.

Q87

If a source of power

4kW

produces

{10^{20}}

photons/second, the radiation belongs to a part of the spectrum called

A

X

-rays

B ultraviolet rays

C microwaves

D

\gamma

- rays

Correct Answer

Option A

Solution

Power,

P = {{nhv} \over t}

\Rightarrow v = {{P \times t} \over {nh}}

= {{4 \times {{10}^3} \times 1} \over {{{10}^{20}} \times 6.63 \times {{10}^{ - 34}}}} = 6 \times {10^{16}}Hz

Q88

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

: A metallic surface is irradiated by a monochromatic light of frequency

v > {v_0}

(the threshold frequency). The maximum kinetic energy and the stopping potential are

{K_{\max }}

and

{V_0}

respectively. If the frequency incident on the surface is doubled, both the

{K_{\max }}

anmd

{V_0}

are also doubled. Statement -

2

: The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

A Statement -

1

is true, Statement -

2

is true, Statement -

2

is the correct explanation of Statement -

1

.

B Statement -

1

is true, Statement -

2

is true, Statement -

2

is not the correct explanation of Statement -

1

.

C Statement -

1

is false, Statement -

2

is true.

D Statement -

1

is true, Statement -

2

is false.

Correct Answer

Option C

Solution

By Einstein photoelectric equation,

{K_{\max }} = e{V_0} = hv - h{v_0}

When

v

is doubled,

{K_{\max }}

and

{V_0}

become more than double.

Q89

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths

{\lambda _A}

to

{\lambda _B}

after the collision is:

A

{{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 3}

B

{{{\lambda _A}} \over {{\lambda _B}}} = 2

C

{{{\lambda _A}} \over {{\lambda _B}}} = {2 \over 3}

D

{{{\lambda _A}} \over {{\lambda _B}}} = {1 \over 2}

Correct Answer

Option B

Solution

From question, mA = M; mB =

{m \over 2}

uA = V and uB = 0 Let after collision velocity of A = V1 and velocity of B = V2 Applying law of conservation of momentum, mu = mv1 +

\left( {{m \over 2}} \right){v_2}

$\Rightarrow$ 24= 2v1 + v2 ........(i) By law of collision

e = {{{v_2} - {v_1}} \over {u - 0}}

$\Rightarrow$ u = v2 - v1 ..........(ii) [As collision is elastic, e = 1] using eqns (i) and (ii) v1 =

{4 \over 3}

and v2 =

{4 \over 3}v

We know, de-Broglie wavelength $\lambda$ =

{h \over p}

$\therefore$

{{{\lambda _A}} \over {{\lambda _B}}} = {{{P_B}} \over {{P_A}}}

=

{{{m \over 2} \times {4 \over 3}u} \over {m \times {4 \over 3}}}

= 2

Q90

If the de Broglie wavelengths associated with a proton and an

\alpha

-particle are equal, then the ratio of velocities of the proton and the

\alpha

-particle will be :

A 4 : 1

B 2 : 1

C 1 : 2

D 1 : 4

Correct Answer

Option A

Solution

We know,

{\lambda _p} = {h \over {{p_p}}}

=

{h \over {{m_p}{v_p}}}

Similarly,

{\lambda _\alpha } = {h \over {{m_\alpha }{v_\alpha }}}

Given,

{\lambda _p} = {\lambda _\alpha }

$\Rightarrow$

{h \over {{m_p}{v_p}}} = {h \over {{m_\alpha }{v_\alpha }}}

$\therefore$

{{{v_p}} \over {{v_\alpha }}} = {{{m_\alpha }} \over {{m_p}}}

=

{{4{m_p}} \over {{m_p}}} = {4 \over 1}