Tag: photons

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Questions Related to photons

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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

  1. Is zero for all

  2. Is same for all

  3. Lies between zero and infinity

  4. Lies between zero and a finite maximum
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $v$.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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 

  1. 0.15 e V

  2. 1.5 e V

  3. 15 e V

  4. 150 e V
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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

  1. Vary with intensity of light

  2. Vary with frequency of light

  3. Vary with speed of light

  4. Vary irregulary
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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

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

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

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

  4. $3.2\times 10^{-15}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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

  1. $\sqrt2 :1$

  2. $2\sqrt2 :1$

  3. 2 :1

  4. 4 : 1
Reveal answer Fill a bubble to check yourself
C Correct answer
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,

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

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

$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

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

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

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

  4. $K _{2}=3K _{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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,$k _{1}< \dfrac{k _{2}}{3}$

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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 ($h = 6.4 \times 10^{-34} Js$)

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

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

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

  4. $1.4\times 10^{-19}eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
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).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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

  1. 5000 $A^{0}$

  2. 3100 $A^{0}$

  3. 1700 $A^{0}$

  4. 2700 $A^{0}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$E=\dfrac{hc}{\lambda}$

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

        $\lambda=2700{A}^{0}$
Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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 :

  1. 8 eV

  2. 6 eV

  3. 2 eV

  4. 4 eV
Reveal answer Fill a bubble to check yourself
D Correct answer
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).

Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

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.

  1. $2.34 \ eV$

  2. $0.85 \ eV$

  3. $1.13 \ eV$

  4. $3.32 \ eV$
Reveal answer Fill a bubble to check yourself
C Correct answer
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).