Tag: physics

Questions Related to physics

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. $2800\ m/s$

Show answer
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. $3.2 \times 10^{-15}$

Show answer
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

Show answer
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

Show answer
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 } } $

Show answer
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

A hydrogen-like atom is in a higher energy level of quantum number $6$. The excited atom make a transition to first excited state by emitting photons of total energy $27.2\ eV$. The atom from the same excited state make a transition to the second excited state by successively emitting two photons. If the energy of one photon is $4.25\ eV$, find the energy of other photon.

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $5.25\ eV$

  2. $6.25\ eV$

  3. $6.95\ eV$

  4. $7.80\ eV$

Show answer
Correct Option: A
Explanation:
Total energy liberated during transition of ${ e }^{ - }$ from ${ n }^{ th }$ shell to first excited state
i.e, ${ 2 }^{ nd }$ shell $=10.20+17.0=2720eV$
                     $=27.20\times 1.602\times { 10 }^{ -12 }erg$
$\dfrac { hc }{ \lambda  } ={ R } _{ H }\times { Z }^{ 2 }{ h } _{ c }\left[ \dfrac { 1 }{ { z }^{ 2 } } -\dfrac { 1 }{ { n }^{ 2 } }  \right] $
$27.20\times 1.602\times { 10 }^{ -12 }={ R } _{ H }\times { Z }^{ 2 }{ h } _{ c }\left[ \dfrac { 1 }{ { z }^{ 2 } } -\dfrac { 1 }{ { n }^{ 2 } }  \right] \quad \longrightarrow \left( 1 \right) $
i.e. ${ 3 }^{ rd }$ shell $=4.25+5.95=10.20eV$
                     $=10.20\times 1.602\times { 10 }^{ -12 }erg$
$\therefore$   $10.20\times 1.602\times { 10 }^{ -12 }={ R } _{ H }\times { Z }^{ 2 }{ h } _{ c }\left[ \dfrac { 1 }{ { 3 }^{ 2 } } -\dfrac { 1 }{ { n }^{ 2 } }  \right] $ $\longrightarrow \left( 2 \right) $
We get $n=5.25eV$

Calculate the number of photons emitted per seconds from a monochromatic light source of $40\ W$, giving out light of wavelength $5000\ \mathring {A}$.

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $1.0\ \times  10^{30}$

  2. $1.0\ \times  10^{20}$

  3. $1.0\ \times  10^{10}$

  4. $1.0\ \times  10^{25}$

Show answer
Correct Option: B

In a photoelectric cell, illuminated with a certain radiation, the minimum negative anode of potential with respect to emitting metal required to stop the electron is $2 V.$ the  minimum KE of the photoelectrons is 

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $0 eV$

  2. $1 eV$

  3. $2 eV$

  4. $ 4 eV$

Show answer
Correct Option: C
Explanation:

Given,

$V _0=2V$
The minimum kinetic energy of the photo electron is
$K _{min}=eV _0$
$K _{min}=2eV$
The correct option is C.