Tag: physics

Questions Related to physics

The difference in the electron energies associated wiwth the two state of an atom is $4 eV$. if $\frac { h }{ e } =4\times { 10 }^{ -5 }{ JsC }^{ -1 }$, the wavelength of the photon emitted as a result of the above transition will be

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $6000 A$

  2. $3000 A$

  3. $1000 A$

  4. $9000 A$

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Correct Option: A

When a hydrogen atom emits a photon of energy $12.09eV$,it's orbit's angular momentum changes by (where $h$ is Planck's constant) ?

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\dfrac{3h}{\pi}$

  2. $\dfrac{2h}{\pi}$

  3. $\dfrac{h}{\pi}$

  4. $\dfrac{4h}{\pi}$

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Correct Option: C
Explanation:

Given that,

Emission of photon of 12.1eV corresponds to the transition from

$n=3,n=1$

Now, change in angular momentum

  $ =\left( {{n} _{2}}-{{n} _{1}} \right)\times \dfrac{h}{2\pi } $

 $ =\left( 3-1 \right)\times \dfrac{h}{2\pi } $

 $ =\dfrac{h}{\pi } $

Hence, this is the required solution 

Assuming photo-emission to take place, the factor by which the maximum velocity of the emitted photo electrons changes when the wavelength of the incident radiation is increased four times, is (assuming work function to be negligible in comparison to $hcl\lambda $)

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 4

  2. $\dfrac{1}{4}$

  3. 2

  4. $\dfrac{1}{2}$

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Correct Option: D
Explanation:

Einstein's Photoelectric equation is given as : $\dfrac{hc}{\lambda } = \phi + \dfrac{1}{2} mv^{2}$ (where $\phi $ = work function , v = velocity of photoelectron, $\lambda $ = wavelength of incident radiation) 

 as $\dfrac{hc}{\lambda } >>\phi \Rightarrow \dfrac{hc}{\lambda }\approx \dfrac{1}{2} mv^{2}$

 now wavelength is increased by 4 times : 

 $\Rightarrow \lambda _{2}=4\lambda $ 

 $\Rightarrow \dfrac{hc}{4\lambda }=\dfrac{1}{2} mv _{2}^{2}$

 $\Rightarrow \dfrac{1}{4}\left ( \dfrac{1}{2} mv^{2}\right )=\dfrac{1}{2}mv _{2}^{2}$ 

 $v _{2}^{2}=\dfrac{v^{2}}{4}\Rightarrow v _{2}=\dfrac{v}{2}$ 

 so maximum velocity of photoelectrons will be $\dfrac{1}{2}$ times when wavelength becomes 4 times.

Momentum of $\gamma-ray$ photon of energy $3\ keV$ in $kg-m/s$ will be

physics quantum physics photons concept of photon photons and photoelectric effect
  1. $2.95 \times 10^{-23}$

  2. $1.6\times 10^{-21}$

  3. $1.6\times 10^{-24}$

  4. $1.6\times 10^{-27}$

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Correct Option: A
Explanation:

If the energy of the electron is $E=3KeV=3\times 10^3 eV=3\times 1.6\times 10^{-16}Joule$

 
Then the momentum ,$p$ is given by  $p=\sqrt[2]{2mE}=\sqrt[2]{2\times 9.1\times 10^{-31} \times 4.8\times 10^{-16}}=2.95\times 10^{-23} Kgm/s$
Where $m=9.1\times 10^{-31}Kg$ is mass of electron.

Radiational wave length $ \lambda $=124 nm falls on a metallic surface. Then the kinetic energy of the ejected photo electron(s) can be : (Given that threshold wavelength ($ \lambda _{0} $)=248 nm)

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 1 eV

  2. 2 eV

  3. 3 eV

  4. 5 eV

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Correct Option: D
Explanation:
We know,
$KE=hv-h{ v } _{ 0 }$
        $=hc\left[ \dfrac { 1 }{ \lambda  } -\dfrac { 1 }{ { \lambda  } _{ 0 } }  \right] $
$=6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 }\times \left[ \dfrac { 1 }{ 124\times { 10 }^{ -9 } } -\dfrac { 1 }{ 248\times { 10 }^{ -9 } }  \right] $
$=0.08015\times { 10 }^{ -17 }$
$=8.015\times { 10 }^{ -19 }J$
$1.6\times { 10 }^{ -19 }J=1eV$
$8.015\times { 10 }^{ -19 }J=\dfrac { 1 }{ 1.6\times { 10 }^{ -19 } } \times 8.015\times { 10 }^{ -19 }$
                            $=4.74eV$
Thus, the answer is close to $5eV$.

Total energy of $electron$ is more than energy of $photon$ if both are having $equal\ \lambda.$

physics quantum physics photons concept of photon photons and photoelectric effect

  1. True

  2. False

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Correct Option: A

Monochromatic light of wavelength 440 nm is produced. The power emitted by light is 18 mW, The number of photons emitted per second by light beam is :

$(h=6.6\times { 10 }^{ -34 })$

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $2.09\times { 10 }^{ 16 }$

  2. $4\times { 10 }^{ 16 }$

  3. $3\times { 10 }^{ 18 }$

  4. $4\times { 10 }^{ 18 }$

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Correct Option: B
Explanation:

Given,


$\lambda=440nm$


$h=6.6\times 10^{-34}$

$I=18m W$

The energy of monochromatic light,

$E=\dfrac{hc}{\lambda}$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^8}{440\times 10^{-9}}$

$E=0.045\times 10^{-17}J$

The number of photon emitted per second by the light beam,

$n=\dfrac{I}{E}$ 

$n=\dfrac{18\times 10^{-3}}{0.045\times 10^{-17}}$

$n=4\times 10^{16}$

The correct option is B.

An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\dfrac{25m}{24hR}$

  2. $\dfrac{24m}{25hR}$

  3. $\dfrac{24hR}{25m}$

  4. $\dfrac{25hR}{24m}$

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Correct Option: C

Violet light is falling on a photosensitive material causing ejection of photoelectrons with maximum kinetic energy of $1$ eV. Red light falling on metal will cause emission of photoelectrons with maximum kinetic energy (approximately) equal to

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $1.2$ eV

  2. $0.9$ eV

  3. $0.5$ eV

  4. Zero, that is no photoemision

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Correct Option: D

When an electron de-exited back from ${\left( {n + 1} \right)^{th}}$ state to ${n^{th}}$ state in a hydrogen like atoms, wavelength of radiations emitted is ${\lambda _1}\left( {n >  > 1} \right)$. In the same atom de-broglies wavelength associated with an electron in $nth$ state is ${\lambda _2}$. Then $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ is proportional to 

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\frac{1}{n}$

  2. n

  3. ${n^2}$

  4. ${n^3}$

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Correct Option: A