Tag: quantum physics

Questions Related to quantum physics

If the work function of the metal is W and the frequency of the incident light is v , then there is no emission of photo-electrons if

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $v< W/h$

  2. $v > W/h$

  3. $v \geq W/h$

  4. $v \leq W/h$

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Correct Option: A
Explanation:

Maximum Kinetic Energy $=h\nu -\phi $
$=h\nu -W $
Since there is no emission
So, $h\nu -W < 0$
or $\nu < \dfrac{W}{h}$

A proton, an electron and an $\alpha$ particle is accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to the magnetic field $\vec{B}$. The ratio of their kinetic energies is

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 2:1:1

  2. 2:2;1

  3. 1:2:1

  4. 1:1:2

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Correct Option: D
Explanation:

Kinetic energy = qV where q is charge and V is potential
But potential is same
So, $K. E \propto q$
$q _p = q _e = 2q$
Putting these values,
we get the ratio of 1:1 :2

Identify which of the following statement is true about the momentum of a photon?

physics quantum physics photons concept of photon photons and photoelectric effect

  1. It is proportional to the wavelength of the photon

  2. It is inversely proportional to the wavelength of the photon.

  3. It is inversely proportional to the square of the wavelength of the photon.

  4. It is proportional to the mass of the photon.

  5. It is equal to the energy of the photon.

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Correct Option: B
Explanation:

Momentum of the photon $p = \dfrac{h}{\lambda}$ where $\lambda$ is the wavelength of the photon

$\implies$$p  \propto \dfrac{1}{\lambda}$
Hence momentum of the photon is inversely proportional to the wavelength of the photon.

What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength $5200\overset {\circ}{A}$?

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $700\ m/s$

  2. $1000\ m/s$

  3. $1400\ m/s$

  4. $2800\ m/s$

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Correct Option: C
Explanation:

Momentum, $p = mv = \dfrac {h}{\lambda}$


or $v = \dfrac {h}{m\lambda}$

$\therefore v = \dfrac {6.62\times 10^{-34}}{9.1\times 10^{-31}\times 5.2\times 10^{-7}}$

$\Rightarrow v = \dfrac {6.2\times 10^{4}}{9.1\times 5.1}$

$\Rightarrow v = 1400\ m/s$.

Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy (in joule) of the fastest electron emitted is :

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $3.2 \times 10^{-21}$

  2. $1.6 \times 10^{-17}$

  3. $3.2 \times 10^{-19}$

  4. $3.2 \times 10^{-15}$

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Correct Option: C
Explanation:

$Energy\quad of\quad radiation,\quad h\nu =6.2eV\ =6.2\times 1.6\times { 10 }^{ -19 }J\ Work\quad function\quad W=4.2eV\ =4.2\times 1.6\times { 10 }^{ -19 }J\ KE=h\nu -W\ =6.2\times 1.6\times { 10 }^{ -19 }-4.2\times 1.6\times { 10 }^{ -19 }\ =2\times 1.6\times { 10 }^{ -19 }\ =3.2{ \times 10 }^{ -19 }J$

Momentum of a photon having frequency $1.5\times 10^{13}Hz$?

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $3.13\times 10^{-29}kg m/s$

  2. $3.3\times 10^{-34}kg m/s$

  3. $6.6\times 10^{-34}kg m/s$

  4. $6.6\times 10^{-30}kg m/s$

Show answer
Correct Option: A
Explanation:

$\quad Momentum\quad of\quad a\quad photon\quad is\quad P=\quad h/\lambda \ \quad And\quad \quad \quad \lambda \nu =c\quad ,\quad where\quad c=\quad speed\quad of\quad light\quad h=\quad Planck's\quad constant\ \qquad \qquad \qquad \qquad \qquad \qquad \lambda =\quad wavelength\quad of\quad photon\ \qquad \qquad \qquad \qquad \qquad \qquad \nu =\quad frequency\quad of\quad photon\ so\quad \lambda =\quad \dfrac { 3\times { 10 }^{ 8 } }{ 1.5\times { 10 }^{ 13 } } =2\times { 10 }^{ -5 }{ m }^{ }\quad P=\dfrac { 6.26\times { 10 }^{ -34 } }{ 2\times { 10 }^{ -5 } } =3.13\times { 10 }^{ -29 }kg m/s\ Therefore\quad option\quad A.\quad$

A beam of light has two wavelengths $4971\mathring{A}$ and $6216\mathring{A}$ with a total intensity of $3.6\times { 10 }^{ -3 }W{ m }^{ -2 }$ equally distributed among the two wavelengths. The beam falls normally on an area of $1{cm}^{2}$ of a clean metallic surface of work function $2.3eV$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo electrons liberated in $2s$ approximately :

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $6\times { 10 }^{ 11 }$

  2. $9\times { 10 }^{ 11 }$

  3. $11\times { 10 }^{ 11 }$

  4. $15\times { 10 }^{ 11 }$

Show answer
Correct Option: B
Explanation:
${ E } _{ 1 }=\cfrac { 1242 }{ 497.2 } =2.50eV;{ E } _{ 2 }=\cfrac { 1242 }{ 6621. } =2.0eV$
so, photoelectron emission takes place only due to first wavelength
$\therefore$ No. of photoelectrons emitted $=\cfrac { 1.8\times { 10 }^{ -3 }\times { 10 }^{ -4 }\times 2 }{ 2.5\times 1.6\times 10^{-19} } hv=9\times { 10 }^{ 11 }$

The threshold frequency for a metallic surface corresponds to an energy of $6.2eV$, and the stopping potential for a radiation incident on this surface $5V$. The incident radiation lies in.

physics quantum physics photons concept of photon photons and photoelectric effect

  1. X-ray

  2. ultra-violet region

  3. infra-red region

  4. visible region

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Correct Option: B
Explanation:

$hv=5eV+6.2eV=11.2eV$
$\lambda =\cfrac { 1242 }{ 11.2 } nm=1109\mathring { A } $
it lies in ultraviolet region

When photon of the energy 3.8 eV falls on metallic surface of work function 2.8 eV, then the kinetic energy of emitted electrons are

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 1 eV

  2. 6.6 eV

  3. 0 to 1 eV

  4. 2.8 eV

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Correct Option: A
Explanation:

Photoelectric law

$KE=h\nu -W\ h\nu =3.8eV\ W=2.8eV\ KE=3.8-2.8\ =1eV$

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference of $V$ volt in vacuum. Its final speed will be

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\cfrac { eV }{ 2m } $

  2. $\cfrac { eV }{ m } $

  3. $\sqrt { \cfrac { 2eV }{ m } } $

  4. $\sqrt { \cfrac { eV }{ 2m } } $

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Correct Option: C
Explanation:
  • Energy gained by electron when accelerated through a potential $V $ is $ charge\times voltage$ i.e., $eV$ 
  • As Kinetic energy 
  •  of any mass $m$ is given as $\dfrac{mv^2}{2}$ 
  • so velocity $v$= $\sqrt[2]{ \dfrac{2K.E.}{m}}$ 
  • for final speed put $K.E.= eV$ we get final speed $v$= $ \sqrt[2]{ \dfrac {2eV}{m}}$. Option C is correct