Tag: quantum physics

Questions Related to quantum physics

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. $7.80\ eV$

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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. $1.0\ \times  10^{25}$

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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.

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

Show answer
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. $\dfrac{1}{2}$

Show answer
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. $1.6\times 10^{-27}$

Show answer
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

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

Show answer
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.