Dual Nature of Radiation and Matter — Page 5 | JEE Physics

Q41

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: (

{{{hc} \over e}}

= 1240 nm eV)

A 0.5 V

B 1.0 V

C 2.0 V

D 1.5 V

Correct Answer

Option B

Solution

{{hc} \over {{\lambda _1}}} = \phi + e

V1 . . . (i)

{{hc} \over {{\lambda _2}}} = \phi + e

V2 . . . (ii) (i) $-$ (ii) hc

\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)

= e(V1 $-$ V2) $\Rightarrow$ V1 $-$ V2 =

{{hc} \over e}

\left( {{{{\lambda _2} - {\lambda _1}} \over {{\lambda _1} - {\lambda _2}}}} \right)

= (1240nm $-$ V)

{{100nm} \over {300nm \times 400nm}}

= 1V

Q42

An electron (mass m) with initial velocity

\overrightarrow v = {v_0}\widehat i + {v_0}\widehat j

is in an electric field

\overrightarrow E = - {E_0}\widehat k

. If

\lambda _0

is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :

A

{{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}

B

{{{\lambda _0}\sqrt 2 } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}

C

{{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {2{m^2}v_0^2}}} }}

D

{{{\lambda _0}} \over {\sqrt {2 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}

Correct Answer

Option C

Solution

\overrightarrow v = {v_0}\widehat i + {v_0}\widehat j

$\lambda$ 0 =

{h \over {m\sqrt 2 {v_0}}}

....(1)

\overrightarrow E = - {E_0}\widehat k

\overrightarrow F = q\overrightarrow E

= (-e)(

- {E_0}\widehat k

) =

e{E_0}\widehat k

$\therefore$

\overrightarrow a = {{\overrightarrow F } \over m}

=

{{e{E_0}\widehat k} \over m}

Velocity at time t,

\overrightarrow {{v_f}}

=

{v_0}\widehat i + {v_0}\widehat j

+

{{e{E_0}} \over m}t\widehat k

\left| {\overrightarrow {{v_f}} } \right|

=

\sqrt {v_0^2 + v_0^2 + {{\left( {{{e{E_0}t} \over m}} \right)}^2}}

=

\sqrt {2v_0^2 + {{{e^2}E_0^2} \over {{m^2}}}{t^2}}

$\therefore$ Wavelength at time t $\lambda$ =

{h \over {m{v_f}}}

=

{h \over {m\sqrt {2v_0^2 + {{{e^2}E_0^2} \over {{m^2}}}{t^2}} }}

=

{h \over {\sqrt 2 m{v_0}\sqrt {1 + {{{e^2}E_0^2} \over {2{m^2}v_0^2}}{t^2}} }}

=

{{{\lambda _0}} \over {\sqrt {1 + {{{e^2}E_0^2} \over {2{m^2}v_0^2}}{t^2}} }}

Q43

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of

\lambda

. An alpha particle having certain kinetic energy has the same de-Brogle wavelength

\lambda

. The ratio of kinetic energy of proton and that of alpha particle is:

A 1 : 4

B 2 : 1

C 4 : 1

D 1 : 2

Correct Answer

Option C

Solution

For same $\lambda_{1}$ momentum should be same, $(P)_{P}=(P)_{\alpha}$ $\Rightarrow \sqrt{2 k_{P} m_{P}}=\sqrt{2 k_{\alpha} m_{\alpha}}$ $\Rightarrow k_{P} m_{P}=k_{\alpha} m_{\alpha}$ $\Rightarrow \dfrac{k_{P}}{k_{\alpha}}=\left(\dfrac{m_{\alpha}}{m_{P}}\right)=\dfrac{4}{1}=4: 1$

Q44

A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :

A Kp < Ke and Pp = Pe

B Kp = Ke and Pp = Pe

C Kp < Ke an Pp < Pe

D Kp > Ke and Pp = Pe

Correct Answer

Option A

Solution

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

{\lambda _e} = {h \over {{P_e}}}

$\because$

{\lambda _p} = {\lambda _e}

\Rightarrow {P_p} = {P_e}

{(K)_p} = {{P_p^2} \over {2{m_p}}}

{(K)_e} = {{P_e^2} \over {2{m_e}}}

Kp < Ke as mp > me Option (a)

Q45

The work functions of Aluminium and Gold are

4.1 ~\mathrm{eV}

and and

5.1 ~\mathrm{eV}

respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

A 1.5

B 1.24

C 1

D 2

Correct Answer

Option C

Solution

We are given the work functions of Aluminium and Gold as

4.1 ~\mathrm{eV}

and

5.1 ~\mathrm{eV}

, respectively. The stopping potential (

V_s

) is related to the frequency (

f

) of the incident light by the equation:

eV_s = h(f - f_0)

Where

e

is the charge of an electron,

h

is Planck's constant, and

f_0

is the threshold frequency. The threshold frequency is related to the work function ( $\phi$ ) by:

\phi = hf_0

So, we can write the equation for stopping potential as:

V_s = \frac{h}{e}(f - \frac{\phi}{h})

This equation represents a straight line with slope

\frac{h}{e}

.

Therefore, the slope of the stopping potential versus frequency plot is the same for both Aluminium and Gold.

The ratio of the slopes is:

\frac{\text{Slope for Gold}}{\text{Slope for Aluminium}} = \frac{\frac{h}{e}}{\frac{h}{e}} = 1

So , the ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is 1.

Q46

When UV light of wavelength

300 \mathrm{~nm}

is incident on the metal surface having work function

2.13 \mathrm{~eV}

, electron emission takes place. The stopping potential is : (Given hc

=1240 \mathrm{~eV} \mathrm{~nm}

)

A 4 V

B 2 V

C 4.1 V

D 1.5 V

Correct Answer

Option B

Solution

To find the stopping potential ( $V_s$ ) when UV light of wavelength $300$ nm is incident on a metal surface with a work function of $2.13$ eV, we can use the photoelectric equation which relates the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted electrons.

The energy (E) of the photons can be calculated using the equation: $E = \dfrac{hc}{\lambda}$ where $h$ is the Planck constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light.

Given that $hc = 1240$ eV nm, we can calculate the energy of the UV light photons directly.

Substituting $hc = 1240$ eV nm and $\lambda = 300$ nm into the equation gives: $E = \dfrac{1240 \, \text{eV nm}}{300 \, \text{nm}} = 4.13 \, \text{eV}$ Next, we can calculate the maximum kinetic energy of the emitted electrons using the photoelectric effect equation: $K_{max} = E - \phi$ where $K_{max}$ is the maximum kinetic energy of the emitted electrons, $E$ is the energy of the incident photons, and $\phi$ is the work function of the metal.

Given $E = 4.13$ eV and the work function $\phi = 2.13$ eV, we have: $K_{max} = 4.13 \, \text{eV} - 2.13 \, \text{eV} = 2 \, \text{eV}$ The stopping potential ( $V_s$ ) is related to the maximum kinetic energy of the emitted electrons by the equation: $K_{max} = eV_s$ where $e$ is the elementary charge (the charge of an electron), and $V_s$ is the stopping potential.

Since $e = 1$ when using energy in eV and potential in volts, the stopping potential $V_s$ can be directly equated to the kinetic energy in eV: $V_s = K_{max} = 2 \, \text{V}$ Therefore, the stopping potential is $2$ V, which corresponds to Option B.

Q47

An

\alpha

particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are

\lambda

\alpha

and

\lambda

p respectively. The ratio

{{{{\lambda _p}} \over {{\lambda _\alpha }}}}

is :

A 8

B 2.8

C 7.8

D 3.8

Correct Answer

Option B

Solution

We know,

qv = {{{p^2}} \over {2m}}

\Rightarrow p = \sqrt {2mqv}

$\therefore$

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

$\therefore$

{{{\lambda _p}} \over {{\lambda _\alpha }}} = {{\sqrt {2{m_\alpha }{q_\alpha }v} } \over {\sqrt {2{m_p}{q_p}v} }}

= {{\sqrt {2(4m)(2e)} } \over {\sqrt {2me} }}

= \sqrt 8

= 2\sqrt 2

Q48

The work function of a substance is

4.0

eV.

The longest wavelength of light that can cause photo-electron emission from this substance is approximately.

A

310

nm

B

400

nm

C

540

nm

D

220

nm

Correct Answer

Option A

Solution

For the longest wavelength to emit photo electron

{{hc} \over \lambda } = \phi \Rightarrow \lambda = {{hc} \over \phi }

\Rightarrow \lambda = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {40 \times 1.6 \times {{10}^{ - 16}}}} = 310nm

Q49

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (mp = 1.00727u me = 0.00055u)

A 41.4 : 1

B (1860)2 : 1

C 1860 : 1

D 43 : 1

Correct Answer

Option D

Solution

{\lambda _e} = {{12.27} \over {\sqrt V }}\mathop A\limits^o

{\lambda _p} = {{0.286} \over {\sqrt V }}\mathop A\limits^o

{{{\lambda _e}} \over {{\lambda _p}}} = {{12.27} \over {0.286}} = 43

Q50

The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is : [me = mass of electron = 9

\times

10

-

31 kg, h = Planck constant = 6.6

\times

6.6

\times

10

-

34 Js, kB = Boltzmann constant = 1.38

\times

10

-

23 JK

-

1]

A 6.26 nm

B 8.46 nm

C 2.26 nm

D 3.25 nm

Correct Answer

Option A

Solution

Given, Planck's constant, h = 6.6 $\times$ 10 $-$ 34 Js Boltzmann constant, kB = 1.38 $\times$ 10 $-$ 23 J/K Mass of an electron, me = 9 $\times$ 10 $-$ 31 kg Temperature of an ideal gas, T = 300 K As we know that, de-Broglie wavelength,

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

.... (i) Here, E is the kinetic energy,

E = {{3{K_B}T} \over 2}

Substituting value of E in Eq. (i), we get

\lambda = {h \over {\sqrt {3m{K_B}T} }}

Substituting the given values in the above equation, we get

\lambda = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}

= 6.26 nm $\therefore$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.