Tag: quantum physics

Questions Related to quantum physics

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 1:1

  2. 2:1

  3. 1:2

  4. 1:4

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Correct Option: A

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\cfrac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case, the work function of the surface of the material is  ($h=$ Planks constant, $c=$ speed of light)

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\cfrac{hc}{3\lambda}$

  2. $\cfrac{hc}{2\lambda}$

  3. $\cfrac{hc}{\lambda}$

  4. $\cfrac{2hc}{\lambda}$

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Correct Option: B

In photoelectric effect for silver threshold is $\lambda _0 = 3250 \times 10^{-10} m$. If U.V of $\lambda = 2536 \times 10^{-10}$ is incident then velocity of electron from will be 

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $6 \times 10^6$

  2. $3 \times 10^6$

  3. $6 \times 10^5$

  4. $3 \times 10^5$

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Correct Option: C

Find the maximum $KE$ of photoelectrons emitted from the surface of lithium$(\phi=2.39 eV)$ when exposed with $\displaystyle E=E _{0}(1+\cos 6\times10^{14}t)\cos 3.6\times 10^{15}t$

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 0.37 $eV$

  2. 0.1 $eV$

  3. 0.02 $eV$

  4. 0.06 $eV$

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Correct Option: C
Explanation:

For maximum $KE.$ frequency should be maximum. Here,
$\displaystyle w=3.6\times10^{15}s^{-1}$


$\displaystyle (KE) _{max}=\dfrac{hw}{2\pi}-\phi$

$=\displaystyle =\dfrac{6.625\times10^{-34}\times3.6\times10^{+15}}{6.28\times1.6\times10^{-19}}-2.39$

$=2.36-2.39$
$=0.02$ $eV$
So, the answer is option (C).

Two separate monochromatic light beams A and B of the same intensity are falling normally on a unit area of a metallic surface. Their wavelengths are $\lambda _A$ and $\lambda _B$, respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam A to that from B is

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $(\lambda _A/ \lambda _B)^2$

  2. $\lambda _A/ \lambda _B$

  3. $\lambda _B/ \lambda _A$

  4. 1

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Correct Option: B

The momentum of a photon of energy 1MeV, in kg/m/s, will be :

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\displaystyle 10^{-22}$

  2. $\displaystyle 0.33\times 10^{6}$

  3. $\displaystyle 5\times 10^{-22}$

  4. $\displaystyle 7\times 10^{-24}$

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Correct Option: C
Explanation:

The momentum of photon is given by


$p = \dfrac{E}{c}$

where, E is the energy of photon(in eV) and c is the velocity of light , $c = 3 \times 10^8 ms^{-1}$.

$p = \dfrac{1 \times 10^{6} \times 1.6 \times 10^{-19}}{3 \times 10^8}$


$p =  0.533 \times 10^{-21}$

$p = 5 \times 10^{-22} kgsm^{-1}$

So, the answer is option (C).

When light is made incident on a surface, then photoelectrons are emitted from it. The kinetic energy of photoelectrons

physics quantum physics photons concept of photon photons and photoelectric effect

  1. Depends on the wavelength of incident light

  2. Is same

  3. Is more than a certain minimum value

  4. None of these

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Correct Option: A
Explanation:

The photoelectric effect related with the incident light frequency $(\nu)$ by the following equation:
$E _k=eV _s=h\nu-\phi$,
where, $\phi$ is the work function of the material and $E _k$ is the kinetic energy of the photo electron. So the photoelectrons emitted from the surface of sodium metal are of speeds from zero to a certain maximum depending on the incident photon energy. So, the kinetic energy of photoelectrons depends on the wavelength of incident light.

So, the answer is option (A).

The work function of aluminum is 4.2eV. Light of wavelength $\displaystyle 2000\mathring {A} $ is incident on it. The maximum energy of emitted electrons will be

physics quantum physics photons concept of photon photons and photoelectric effect

  1. Zero

  2. 1.99 volt

  3. 3.2 volt

  4. 6.19 volt

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Correct Option: B
Explanation:

The work function of aluminium is
$\phi = 4.2 eV = 4.2 \times 1.6 \times 10^{-19}$
$\phi = 6.72 \times 10^{-19} J$
The energy of incident photon is

$h\nu = \dfrac{hc}{\lambda}$
$h\nu = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{2000\times 10^{-10}}$
$h\nu = 9.9 \times 10^{-19} J$
The maximum energy of photoelectrons is
$E _{max} = h\nu - \phi$
$E _{max} = 9.9 \times 10^{-19} - 6.72 \times 10^{-19}$
$E _{max} = 3.18 \times 10^{-19} J$
$E _{max} = \dfrac{3.18 \times 10^{-19}}{1.6 \times 10^{-19}}$
$E _{max} = 1.99 eV$
So, the answer is option (B).

The work function of aluminum is 4.2eV. Light of wavelength $\displaystyle 2000\mathring {A} $ is incident on it. The minimum energy of emitted electrons will be 

physics quantum physics photons concept of photon photons and photoelectric effect

  1. Zero

  2. 2.0 eV

  3. 4.2 eV

  4. 6.19 eV

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Correct Option: B
Explanation:

Energy of the incident light is $6.2 eV$. Now the work function of aluminum is $4.2$. So the  minimum energy of emitted electrons will be $(6.2-4.2)=2$ eV.

So, the answer is option (B).

The work function of a metal is X eV. When light of energy 2X is made incident on it then the maximum kinetic energy of emitted photoelectron will be 

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $2.5X \ eV$

  2. $2X \ eV$

  3. $X \ eV$

  4. $3X \ eV$

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Correct Option: C
Explanation:

The maximum kinetic energy of emitted photoelectron is given as
$E _{max} = h\nu - \phi$
where,$\nu$ is the frequency of incident light and $\phi$ is photoelectric work function of metal. 
$E _{max} = 2X - X$ ...................(given)
$E _{max} = X \ eV$

So, the answer is option (C).