Tag: quantum physics

Questions Related to quantum physics

The photoelectrons emitted from a metal surface are such that their velocity

physics quantum physics photons concept of photon photons and photoelectric effect

  1. Is zero for all

  2. Is same for all

  3. Lies between zero and infinity

  4. Lies between zero and a finite maximum

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Correct Option: D
Explanation:

Suppose radiation of energy $E + \phi$ is incident on a metal surface
$\phi$ is the work function and is the energy need to eject a photoelectron..
Assume all of E gets converted to Kinetic energy.


$E = \dfrac{1}{2}mv^{2}$ 

Hence, $v = \sqrt{\dfrac{2E}{m}}$

There would be molecules that absorb energy lying from $0$ to $E$.
And hence their velocities would lie between $0$ and $v$.

The photoelectric threshold of Tungsten is 2300$ \mathring {A }$. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 18000 $ \mathring {A }$ is 

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 0.15 e V

  2. 1.5 e V

  3. 15 e V

  4. 150 e V

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Correct Option: A
Explanation:

$E _k=d\dfrac{h}{c}\left ( \dfrac{1}{\lambda } -\dfrac{1}{\lambda _o}\right )(in \, e\, V)$

$=\dfrac{6.6\times 10^{34}\times 3\times 10^8}{1.6 \times 10^{-19}}$$\left ( \dfrac{10^{10}}{1800}-\dfrac{10^{10}}{2300} \right )=0.15 e V$

When light is incident on a metal surface the maximum kinetic energy of emitted electrons

physics quantum physics photons concept of photon photons and photoelectric effect

  1. Vary with intensity of light

  2. Vary with frequency of light

  3. Vary with speed of light

  4. Vary irregulary

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Correct Option: B
Explanation:

Photoelectric effect
$KE _{max} = hv - W _o$
where,$W _o$ is the work function of the metal
Clearly $KE$ varies linearly with the the frequency of the photons.

Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). The kinetic energy in joule of the faster electron emitted is approximately

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $3.2\times 10^{21}$

  2. $3.2\times 10^{-19}$

  3. $3.2\times 10^{-17}$

  4. $3.2\times 10^{-15}$

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Correct Option: B
Explanation:

$E _k=E-W _o=6.2-4.2=2.0 e V$
$=2.0 \times 1.6 \times 10^{-19}=3.2\times 10^{-19}\, J$

The ratio of de-Broglie wavelengths of proton and $\alpha$-particle having same kinetic energy is

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $\sqrt2 :1$

  2. $2\sqrt2 :1$

  3. 2 :1

  4. 4 : 1

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Correct Option: C
Explanation:

De Broglie wavelength is given by:


$\lambda = \dfrac{h}{p}$ 

Writing momentum as a a function of kinetic energy and mass

$\lambda = \dfrac{h}{\sqrt{2Em}}$

i.e.   $\dfrac{\lambda _{proton}}{\lambda _{alpha}} = \sqrt{\dfrac{m _{alpha}}{m _{proton}}}$

$m _{alpha} = 4m _{proton}$
So,

$\dfrac{\lambda _{proton}}{\lambda _{alpha}} = \sqrt{\dfrac{m _{alpha}}{m _{proton}}} = \sqrt{\dfrac{4}{1}} = 2$

$K _{1} $and $K _{2}$ are the maximum kinetic energies of the photoelectrons emitted when light of wavelength $\lambda _{1} $ and $\lambda _{2} $  respectively are incident on a metallic surface. If $\lambda _{1}= $3$\lambda _{2} $  then

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $K _{1}>\dfrac{K _{2}}{3}$

  2. $K _{1}<\dfrac{K _{2}}{3}$

  3. $K _{1}=3K _{2}$

  4. $K _{2}=3K _{1}$

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Correct Option: B
Explanation:

$K _{1}=\dfrac{hc}{\lambda _{1} }-\phi $-----------(1)

$K _{2}=\dfrac{hc}{\lambda _{2}}-\phi$-----------(2)

$\because\lambda _{1} = 3 \lambda _{2}$

$k _{2}=3\left(\dfrac{hc}{\lambda _{1}}\right)-\phi$---------(3)

$\dfrac{k _{2}}{3}=\dfrac{hc}{\lambda _{1}}-\dfrac{\phi}{3}$----------(4)

$\dfrac{k _{2}}{3}=(k _{1}+\phi)-\dfrac{\phi}{3}$  (by (1))

$\dfrac{k _{2}}{3}=k _{1}+\dfrac{2 \phi}{3}$

So,$k _{1}< \dfrac{k _{2}}{3}$

The work function of a metal is $1.6\times 10^{-19}$J. When the metal surface is illuminated by the light of wavelength 6400 $A^{o}$, then the maximum kinetic energy of emitted photoelectrons will be ($h = 6.4 \times 10^{-34} Js$)

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $14\times 10^{-19}J$

  2. $2.8\times 10^{-19}J$

  3. $1.4\times 10^{-19}J$

  4. $1.4\times 10^{-19}eV$

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Correct Option: C
Explanation:

$K.E. _{max}=\dfrac{hc}{\lambda }-\phi $


$=\dfrac{6.4\times 10^{-34}\times 3\times 10^{8}}{6.4\times 10^{-7}}-1.6\times 10^{-19}$

$=3\times 10^{-19}-1.6\times 10^{-19}$
$=1.4\times 10^{-19}J.$
So, the answer is option (C).

The work function of a metal is 4.6eV. The wavelength of incident light required to emit photo-electrons of zero energy from its surface, will be

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 5000 $A^{0}$

  2. 3100 $A^{0}$

  3. 1700 $A^{0}$

  4. 2700 $A^{0}$

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Correct Option: D
Explanation:
$E=\dfrac{hc}{\lambda}$

$ 4.6eV=\dfrac{1240eV}{\lambda}$

        $\lambda=2700{A}^{0}$

The photoelectric work function of a metal surface is 2eV. When light of frequency $1.5 \times10^{15}$ Hz is incident on it, maximum kinetic energy of the photo-electrons, approximately is :

physics quantum physics photons concept of photon photons and photoelectric effect

  1. 8 eV

  2. 6 eV

  3. 2 eV

  4. 4 eV

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Correct Option: D
Explanation:

$K.E. _{max}=h\nu-\phi $


=$\dfrac{6.6\times 10^{-34}\times 1.5\times 10^{15}}{1.6\times 10^{-19}}eV-2eV$

$=6eV-2eV$

$=4eV.$

So, the answer is option (D).

Work function of a metal is 3.0eV. It is illuminated by a light of wavelength $3 \times 10^{-7}$m. Then the maximum energy of the electron is.

physics quantum physics photons concept of photon photons and photoelectric effect

  1. $2.34 \ eV$

  2. $0.85 \ eV$

  3. $1.13 \ eV$

  4. $3.32 \ eV$

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Correct Option: C
Explanation:

$Maximum \ \ energy =\dfrac{hc}{\lambda}-\phi  (3\times 10^{-7}m=300nm)$

$=\dfrac{1240}{300} - 3 \ \ \ \ (hc =  1240 \ eV / X nm)$
$=4.13-3$
$=1.13 eV.$
So, the answer is option (C).