Tag: quantum physics

Questions Related to quantum physics

In photoelectric effect, initially when energy of electrons emitted is $E _{0}$, de-Broglie wavelength associated with them is $\lambda _{0}$. Now, energy is doubled then associated de-Broglie wavelength $\lambda^{'}$ is

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\displaystyle\lambda^{'}=\frac{\lambda _{0}}{\sqrt{2}}$

  2. $\displaystyle\lambda^{'}=\sqrt{2}\lambda _{0}$

  3. $\displaystyle\lambda^{'}=\lambda _{0}$

  4. $\displaystyle\lambda^{'}=\frac{\lambda _{0}}{2}$

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Correct Option: A
Explanation:

de-Brogile wavelength is given by
$\displaystyle\lambda =\frac{h}{p}$, where h= Planck's constant and p= momentum
Also energy (E) and momentum are related as 
$\displaystyle E =\frac{p^{2}}{2m}$
$\displaystyle \therefore p=\sqrt{2mE}$
$\displaystyle \therefore \lambda =\frac{h}{\sqrt{2mE}}\times \frac{1}{\sqrt{E}}$ as h and m are constants
Hence, $\displaystyle \frac{\lambda _{0}}{{\lambda}'}=\sqrt{\frac{{E}'}{E}}=\sqrt{\frac{2E}{E}}=\sqrt{2}$
$\displaystyle \therefore  {\lambda}'= \frac{\lambda _{0}}{\sqrt{2}}$

When the momentum of a photon is changed by an amount $p'$ then the corresponding change in the de-Broglie wavelength is found to be $0.20$%. Then, the original momentum of the photon was

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $300 p'$

  2. $500 p'$

  3. $400 p'$

  4. $100 p'$

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Correct Option: B
Explanation:
As, we know de-Broglie wavelength,
$\lambda = \dfrac{h}{p}$
$\therefore \lambda \propto \dfrac{1}{p}$
$\Rightarrow \dfrac{\Delta p}{p} = - \dfrac{\Delta \lambda}{\lambda} \therefore \left| \dfrac{\Delta p}{p} \right | = \left| \dfrac{\Delta \lambda}{\lambda} \right |$
$\Rightarrow \dfrac{p'}{p} = \dfrac{0.20}{100} = \dfrac{1}{500}$
or, $p = 500 p'$

In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is $1eV$ for incoming radiation of frequency $v _{0}$ and $3eV$ for incoming radiation of frequency $3v _{0}/2$. What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency $9v _{0}/4$?

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $3\ eV$

  2. $4.5\ eV$

  3. $6\ eV$

  4. $9\ eV$

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Correct Option: C
Explanation:

$(KE) _{max} = hv - \phi _{0}$
So, $1\ eV = hv _{0} - \phi _{0} .... (i)$
and $3\ eV = \dfrac {hv _{0}}{2} - \phi _{0} .... (ii)$
$\Rightarrow 3\ eV - 1\ eV = \dfrac {hv _{0}}{2}$
or $hv _{0} = 4\ eV$
From Eq. (i), $\phi _{0} = hv _{0} - 1\ eV$
$= 4\ eV - 1\ eV = 3\ eV$
$\therefore (KE) _{mas} = h\times \dfrac {9v _{0}}{4} - 3\ eV$
$= \dfrac {9}{4} (4\ eV) - 3\ eV = 6\ eV$.

Light of wavelength $\lambda$ strikes a photo sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$, then the wavelength must be changed to $\lambda'$, where :

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\lambda' > \lambda$

  2. $\lambda' = \dfrac {\lambda}{2}$

  3. $\lambda' = 2\lambda$

  4. $\dfrac {\lambda}{2} > \lambda' > \lambda$

Show answer
Correct Option: D
Explanation:

We have, $E _{k} = \dfrac {hc}{\lambda} - \phi _{0}$


and $2E _{k} = \dfrac {hc}{\lambda'} - \phi _{0}$

By the two relations, we have

$\lambda' > \dfrac {\lambda}{2}$

and $\lambda' < \lambda$

So, $\dfrac {\lambda}{2} < \lambda' < \lambda$.

A photo-cell is illuminated by a source of light, which is placed at a distance $d$ from the cell. If the distance become $d/2$, then number of electrons emitted per second will be : 

physics quantum physics photons concept of photon photons and photoelectric effect

  1. Remain same

  2. Four times

  3. Two times

  4. One-fourth

Show answer
Correct Option: D
Explanation:

Intensity, $I=\dfrac { E }{ At } \ $


where E is the total energy of source,


A is area of illuminated surface,

t is time

$\therefore I=\dfrac { E }{ 4\pi { { { r }^{ 2 } }t } } \\ \dfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\dfrac { E }{ 4\pi { { { r } _{ 1 }^{ 2 } }t } } \times \dfrac { 4\pi { { { r } _{ 2 }^{ 2 } }t } }{ E } \\ =\dfrac { { r } _{ 2 }^{ 2 } }{ { r } _{ 1 }^{ 2 } } \\ \dfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\dfrac { 4 }{ 1 } \\ \Rightarrow \dfrac { { I } _{ 2 } }{ { I } _{ 1 } } =\dfrac { 1 }{ 4 } $

Since, $\propto$ number of photoelectrons emitted

Therefore, Number of electrons emitted is a quarter of the initial number.

The formula for kinetic mass of photon is:
where $h$ is Planck's constant and $v,\lambda, c$ are frequency, wavelength and speed of photon respectively.

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\cfrac { hv }{ \lambda } $

  2. $\cfrac { h }{ c\lambda } $

  3. $\cfrac { hv }{ c } $

  4. $\cfrac { h\lambda }{ c } $

Show answer
Correct Option: B
Explanation:

Energy of the photon is given by,

$E=hv=\dfrac{hc}{\lambda}$,  ...(1)
also for particle of zero mass,
$E=mc^2$,   ...(2)
form eq(1) and (2) we get,
$mc^2=\dfrac{hc}{\lambda}$
$\therefore$ kinetic mass of photon$= m=\dfrac{h}{c\lambda}$

If momentum of a photon of an electromagnetic radiation is $3.3\times 10^{-29}$ kg m/sec, then frequency of associated wave is:

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $3.0\times 10^3Hz$

  2. $6.0\times 10^3Hz$

  3. $7.5\times 10^5Hz$

  4. $1.5\times 10^{13}Hz$

Show answer
Correct Option: D
Explanation:

$p=\dfrac {hv}{c}$
$\Rightarrow v=\dfrac {cp}{h}=\dfrac {(3\times 20^8)\times (3\cdot 3\times 10{-29})}{6\cdot 6\times 10^{-34}}$


$=1\cdot 5\times 10^{13}Hz.$

Assume that a neutron breaks into a proton and an electron. The energy released suring this process is (mass of neutron $= 1.6725 \times 10^{-27} kg,$
mass od proton $= 1.6725 \times 10^{-27} kg,$ mass of electron$= 9 \times 10^{-31}kg)$

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 0.73 MeV

  2. 7.10 MeV

  3. 6.30 MeV

  4. 5.4 MeV

Show answer
Correct Option: A
Explanation:

$\triangle m = (m _P + m _e) -m _n$

        $= 9 \times 10^{-31} kg $
 $(mc^2) \space energy \space released = (9 \times 10^{-31} kg) c^2 \space joules$
$= \dfrac{9 \times 10^{-31} \times (3 \times 10^8 )^2}{1.6 \times 10^{-13}} Mev$

$= 0.73 Mev$
Hence option (A) is correct

A monochromatic source of light is placed at a large distance $d$ from a metal surface. Photoelectrons are ejected at rate $n$, the kinetic energy being $E$. If the source is brought nearer to distance $d/2$, the rate and kinetic energy per photoelectron become nearly :

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 2n and 2E

  2. 4n and 4E

  3. 4n and E

  4. n and 4E

Show answer
Correct Option: C
Explanation:

The rate is inversely proportional to square of the distance: $n\propto \dfrac{1}{{r}^{2}}$


So the new rate will be $4n$

Kinetic energy is not related to the distance, hence it will remain same $E$.

The work function of a certain metal is $3.31 \, \times \,  10^{-19} J.$ Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 A is
$(Given\, h  = 6.62\times10^{-34} J-s, \, c= 3\times10^{-8} \, ms^{-1},\, e=  1.6 \times10^{-19} \, C)$

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $248 eV$

  2. $0.41 eV$

  3. $2.07 eV$

  4. $0.82 eV$

Show answer
Correct Option: B
Explanation:

Work function $W _0 \, = \, 3.31 \, \times \, 10^{-19} \, J$
Wavelength of incident radiation
 $\lambda \, = \, 5000 \, \times \, 10^{-10} \, m$
$E \, = \, W _0 \, + \, KE$
(According to Einstein equation)
$\displaystyle \frac{hc}{\lambda} \, 3.31 \, \times \, 10^{-19} \, + \, KE$
$KE \, = \, - \, 3.31 \, \times \, 10^{-19} + \, \displaystyle \frac{6.62 \, \times \, 10^{-34} \, \times \, 3 \, \times \, 10^8}{5000 \, \times \, 10^{10}}$
$= \, - \, 3.31 \, \times \, 10^{-19} \, + \, 10^{-19} \, + \, \displaystyle \frac{6.62 \, \times \, 3}{5} \, \times \, 10^{-19}$
$= \, (- \, 3.31 \, \times \, 1.324 \, \times \, 3) \, \times \, 10^{-19}$
$= \, (3.972 \, - \, 3.31) \, \times \, 10^{-19} \, = \, 0.662 \, \times \, 10^{-19} \, J$
$\Rightarrow \, E \, = \, \displaystyle \frac{0.662 \, \times \, 10^{19}}{1.6 \, \times \, 10^{-19}} \, = \, 0.41 eV$