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

Q81

Conductivity of a photodiode starts changing only if the wavelength of incident light is less than

660 \mathrm{~nm}

. The band gap of photodiode is found to be

\left(\dfrac{\mathrm{X}}{8}\right) \mathrm{eV}

. The value of

\mathrm{X}

is : (Given,

\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}

)

A 11

B 13

C 15

D 21

Correct Answer

Option C

Solution

To find the value of

X

in the band gap

\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}

, we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy.

The band gap energy (

E_{g}

) of a material is the minimum energy required for an electron to transition from the valence band to the conduction band, thus creating a hole-electron pair and allowing conductivity to occur.

When light with a certain wavelength ( $\lambda$ ) is incident upon the photodiode, if the energy of the photons is greater than or equal to the band gap energy of the photodiode, then the photons can excite electrons and change the conductivity of the photodiode.

The energy of a photon (

E_{photon}

) is given by the equation:

E_{photon} = \frac{hc}{\lambda}

where:

h

is Planck's constant, which is given as

6.6 \times 10^{-34}

J·s.

c

is the speed of light in a vacuum (

\approx 3 \times 10^{8}

m/s). $\lambda$ is the wavelength of the incident light.

Then the band gap energy in terms of electron volts is found by converting the energy from joules to electron volts (eV) using the charge of an electron

e

(

1.6 \times 10^{-19}

C):

E_{g} = \frac{hc}{\lambda e}

Given that the photodiode starts conducting when the wavelength of light is less than

660 \mathrm{~nm}

, we should use this wavelength as the threshold wavelength

\lambda_{threshold}

:

E_{g} = \frac{hc}{\lambda_{threshold}\times e}

E_{g} = \frac{6.6 \times 10^{-34} \text{ J·s} \cdot 3 \times 10^{8} \text{ m/s}}{660 \times 10^{-9} \text{ m} \times 1.6 \times 10^{-19} \text{ C}}

E_{g} = \frac{6.6 \times 3 \times 10^{-26}}{660 \times 1.6} \mathrm{eV}

E_{g} = \frac{19.8}{660 \times 1.6} \mathrm{eV}

E_{g} = \frac{19.8}{1056} \mathrm{eV}

E_{g} = \frac{18.75}{1000} \mathrm{eV}

E_{g} = 1.875 \mathrm{eV}

Now, we have to compare this value to

\left(\frac{\mathrm{X}}{8}\right) \mathrm{eV}

to find the value of

X

:

1.875 = \frac{X}{8}

X = 1.875 \times 8

X = 15

Therefore, the value of

X

is 15. The correct answer is Option C.

Q82

A small mirror of mass

m

is suspended by a massless thread of length

l

. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror (

\mathrm{c}=

speed of light in vacuum and

g=

acceleration due to gravity)

A

\theta=\dfrac{E}{m c \sqrt{g l}}

B

\theta=\dfrac{E}{2 m c \sqrt{g l}}

C

\theta=\dfrac{3 E}{4 m c \sqrt{g l}}

D

\theta=\dfrac{2 E}{m c \sqrt{g l}}

Correct Answer

Option D

Solution

Force due to beam assuming complete reflection $\mathrm{F}=\dfrac{2 \mathrm{P}}{\mathrm{C}}=\dfrac{2}{\mathrm{C}} \dfrac{\mathrm{dE}}{\mathrm{dt}} ; \mathrm{P}$ is power So change in momentum of mirror.

\mathrm{m}(\mathrm{~V}-0)=\int \mathrm{Fdt}=\frac{2}{\mathrm{C}} \int \mathrm{dE}=\frac{2 \mathrm{E}}{\mathrm{C}}

Now using work energy theorem ....... (1)

\begin{aligned} & \mathrm{W}_{\mathrm{g}}=\Delta \mathrm{k} \\ & -\mathrm{mg} \ell(1-\cos \theta)=0-\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{~g} \ell\left(2 \sin ^2 \frac{\theta}{2}\right)=\frac{\mathrm{v}^2}{2} \end{aligned}

as $\theta$ is small

\begin{aligned} & \mathrm{g} \ell 2\left(\frac{\theta}{2}\right)^2=\frac{1}{2} \frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \quad \text{(from eq. (1))}\\ & \mathrm{~g} \ell \theta^2=\frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{~g} \ell}} \end{aligned}

Q83

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy : (Assume

m_{p}=m_{e} \times 1849

)

A 1:43

B 1:62

C 2:43

D 1:30

Correct Answer

Option A

Solution

The de Broglie wavelength (λ) of a particle can be found using the formula:

\lambda = \frac{h}{p}

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

The momentum of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as follows:

p = \sqrt{2mK}

Combining these two equations, we get:

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

Now, we are given that the kinetic energy of the proton and electron is the same.

Let's denote the masses of the proton and electron as

m_p

and

m_e

, respectively. We are given the relationship between the two masses:

m_p = 1849 \times m_e

Let's find the ratio of the de Broglie wavelengths of the proton ( $λ_p$ ) and the electron ( $λ_e$ ):

\frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_pK}}}{\frac{h}{\sqrt{2m_eK}}}

Simplifying the expression, we get:

\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2m_eK}}{\sqrt{2m_pK}}

The 2K terms cancel out:

\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{m_p}}

Substitute the given relationship between the masses:

\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{1849 \times m_e}}

Further simplification:

\frac{\lambda_p}{\lambda_e} = \frac{1}{\sqrt{1849}}

Since 1849 is equal to

43^2

:

\frac{\lambda_p}{\lambda_e} = \frac{1}{43}

Thus, the ratio of the de Broglie wavelengths of the proton and electron having the same kinetic energy is 1:43.

Q84

The electric field at a point associated with a light wave is given by E = 200 [sin (6

\times

1015)t + sin (9

\times

1015)t] Vm

-

1 Given : h = 4.14

\times

10

-

15 eVs If this light falls on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

A 1.90 eV

B 3.27 eV

C 3.60 eV

D 3.42 eV

Correct Answer

Option D

Solution

Frequency of EM waves =

{6 \over {2\pi }} \times {10^{15}}

and

{9 \over {2\pi }} \times {10^{15}}

Energy of one photon of these waves

= \left( {4.14 \times {{10}^{ - 15}} \times {6 \over {2\pi }} \times {{10}^{15}}} \right)

eV and

\left( {4.14 \times {{10}^{ - 15}} \times {9 \over {2\pi }} \times {{10}^{15}}} \right)

eV = 3.95 eV and 5.93 eV $\Rightarrow$ Energy of maximum energetic electrons = 5.93 $-$ 2.50 = 3.43 eV

Q85

The difference between threshold wavelengths for two metal surfaces

\mathrm{A}

and

\mathrm{B}

having work function

\phi_{A}=9 ~\mathrm{eV}

and

\phi_{B}=4 \cdot 5 ~\mathrm{eV}

in

\mathrm{nm}

is:

\{

Given, hc

=1242 ~\mathrm{eV} \mathrm{nm}\}

A 264

B 138

C 540

D 276

Correct Answer

Option B

Solution

Threshold wavelength (

\lambda_{threshold}

) is the maximum wavelength of light required to remove an electron from a metal surface, i.e., to overcome the work function ( $\phi$ ).

The relationship between work function and threshold wavelength is given by:

\phi = \frac{hc}{\lambda_{threshold}}

Where: $\phi$ is the work function in electron-volts (eV)

h

is the Planck's constant

c

is the speed of light

\lambda_{threshold}

is the threshold wavelength In this problem, we are given the product

hc = 1242 ~\mathrm{eV}\,\mathrm{nm}

, and the work functions

\phi_{A} = 9 ~\mathrm{eV}

and

\phi_{B} = 4.5 ~\mathrm{eV}

. Let's calculate the threshold wavelengths for metal surfaces

\mathrm{A}

and

\mathrm{B}

: For metal surface

\mathrm{A}

:

\lambda_{A} = \frac{1242}{\phi_{A}} = \frac{1242}{9}

For metal surface

\mathrm{B}

:

\lambda_{B} = \frac{1242}{\phi_{B}} = \frac{1242}{4.5}

Now let's calculate the difference between the threshold wavelengths (

\Delta\lambda

):

\Delta\lambda = \lambda_{B} - \lambda_{A} = \frac{1242}{4.5} - \frac{1242}{9}

By calculating the difference, we get:

\Delta\lambda = 276 - 138 = 138 ~\mathrm{nm}

So, the difference between the threshold wavelengths for the two metal surfaces is

\boxed{138}\,\mathrm{nm}

.

Q86

The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be :

A 4.34 m/s

B 2.19 m/s

C 3.25 m/s

D 4.17 m/s

Correct Answer

Option D

Solution

(

\Delta

E) Releases when photon going from n = 5 to n = 1

\Delta

E = (13.6 $-$ 0.54) eV = 13.06 eV. Pi = Pf (By linear momentum conservation)

0 = {h \over \lambda } - Mv = {V_{{\mathop{\rm Re}\nolimits} coil}} = {h \over {\lambda M}}

..... (i) &

\Delta E = {{hc} \over \lambda } = {{hc} \over {\lambda M}} \times M = Mc{V_{{\mathop{\rm Re}\nolimits} coil}}

{V_{{\mathop{\rm Re}\nolimits} coil}} = {{\Delta E} \over {Mc}} = {{13.06 \times 1.6 \times {{10}^{ - 19}}} \over {1.67 \times {{10}^{ - 27}} \times 3 \times {{10}^8}}}

= 4.17 m/sec

Q87

Which of the following statement is not true about stopping potential

(\mathrm{V}_0)

?

A It depends upon frequency of the incident light.

B It is

1 / \mathrm{e}

times the maximum kinetic energy of electrons emitted.

C It increases with increase in intensity of the incident light.

D It depends on the nature of emitter material.

Correct Answer

Option C

Solution

Stopping potential is independent of intensity of light. It depends on frequency of light.

Q88

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h = 6.63

\times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}

and

\mathrm{m}_{\mathrm{p}}=1836

times

\mathrm{m}_{\mathrm{e}}

)

A

1: \dfrac{1}{1836}

B

1: \sqrt{1836}

C

1: 1836

D

1: \dfrac{1}{\sqrt{1836}}

Correct Answer

Option C

Solution

To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.

The de-Broglie wavelength $\lambda$ is given by:

\lambda = \frac{h}{p}

where

h

is Planck’s constant and

p

is the momentum. Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:

\lambda_{\text{electron}} = \lambda_{\text{proton}} \implies \frac{h}{p_{\text{e}}} = \frac{h}{p_{\text{p}}} \implies p_{\text{e}} = p_{\text{p}}

Next, the kinetic energy

K

of a particle is related to its momentum

p

and mass

m

by the following formula:

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

Given that the momentum

p

is the same for both the proton and the electron, we can write the kinetic energies as:

K_{\text{e}} = \frac{p^2}{2m_{\text{e}}}

K_{\text{p}} = \frac{p^2}{2m_{\text{p}}}

The ratio of the kinetic energies is therefore:

\frac{K_{\text{e}}}{K_{\text{p}}} = \frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}} = \frac{m_{\text{p}}}{m_{\text{e}}}

Given that the mass of the proton

m_{\text{p}}

is 1836 times the mass of the electron

m_{\text{e}}

, we have:

\frac{m_{\text{p}}}{m_{\text{e}}} = 1836

Thus, the ratio of the kinetic energies is:

\frac{K_{\text{e}}}{K_{\text{p}}} = 1836

Therefore, the correct answer is: Option C

1: 1836

Q89

The work function of a metal is 3 eV . The color of the visible light that is required to cause emission of photoelectrons is

A Red

B Green

C Blue

D Yellow

Correct Answer

Option C

Solution

To determine which color of visible light will cause the emission of photoelectrons from a metal with a work function of 3 eV, we need to consider the relationship between the photon energy, the work function, and the wavelength of light.

The maximum kinetic energy ( $\text{KE}_{\max}$ ) of the emitted photoelectrons is given by: $\text{KE}_{\max} = \dfrac{hc}{\lambda} - \phi$ For photoemission to occur, the energy of the incident photons must be greater than the work function ( $\phi$ ): $\dfrac{hc}{\lambda} > \phi$ This implies: $\lambda Substitute the values$ h = 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} $,$ c = 3 \times 10^{8} \, \text{m/s} $, and$ \phi = 3 \, \text{eV} $:$ \lambda Since the wavelength must be less than 414 nm to cause emission, we need light with a shorter wavelength.

Among the visible colors, blue light has a wavelength in this range, making it the correct choice.

Thus, blue light will cause the emission of photoelectrons with this work function.

Q90

An

\alpha

particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

A

\sqrt2

: 1

B 2

\sqrt2

: 1

C 4

\sqrt2

: 1

D 8 : 1

Correct Answer

Option B

Solution

We know, Momentum

(p) = \sqrt {2m{E_k}}

and

{E_k} = q{V_{acc}}

$\therefore$

p = \sqrt {2mq\,{V_{acc}}}

Both $\alpha$ particle and proton are passed through same potential difference. $\therefore$

{\left( {{V_{acc}}} \right)_\alpha } = {\left( {{V_{acc}}} \right)_p} = v

$\therefore$

{p_\alpha } = \sqrt {2{m_\alpha }{q_\alpha }v}

{p_p} = \sqrt {2{m_p}{q_p}v}

$\therefore$

{{{p_\alpha }} \over {{p_p}}} = \sqrt {{{{m_\alpha }{q_\alpha }} \over {{m_p}{q_p}}}}

= \sqrt {{{4{m_p} \times 2e} \over {{m_p} \times e}}}

= \sqrt {{8 \over 1}}

= {{2\sqrt 2 } \over 1}