Answer

Verified

403.8k+ views

Hint:Einstein explained the photoelectric effect on the basis of Planck’s quantum theory of radiation. According to him, photoelectric emission does not take place by continuous absorption of energy from radiation. Light radiation consists of discrete packets of energy called photons having energy$E = h\nu $

Given: Wavelength of the laser beam,

\[ \Rightarrow \lambda = 6600angstorm\]

$ \Rightarrow 6600 \times {10^{ - 10}}m$

Energy of each photon in the light beam is given by,

$E = h\nu $……………………(1)

Where, $h$ is the Planck’s constant

\[\upsilon \]is the frequency of a laser beam.

We know that, $c = \nu \lambda $

$ \Rightarrow \nu = \dfrac{c}{\lambda }$

Here $C$ is the speed of light.

Equation (1) can also be written as,

$ \Rightarrow E = h\dfrac{c}{\lambda }$

Now substitute the given values, we get

$ \Rightarrow E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6600 \times {{10}^{ - 10}}}}$

$ \Rightarrow E = 3 \times {10^{ - 19}}J$

Power or intensity of a laser beam,

$ \Rightarrow P = 6mW$

$ \Rightarrow 6 \times {10^{ - 3}}W$

Now, the number of photons falling per second on the target is given by,

$ \Rightarrow n = \dfrac{P}{E}$

Substitute the values we get

$ \Rightarrow n = \dfrac{{6 \times {{10}^{ - 3}}W}}{{3 \times {{10}^{ - 19}}J}}$

$\therefore n = 2 \times {10^{16}}$ Photons per second.

Hence, the number of photons passing along path of beam in per second is $2 \times {10^{16}}$

According to Einstein’s photoelectric equation,

When a light is incident on a metal, the photons having energy \[h\nu \] collide with electrons at the surface of the metal. During these collisions, the energy of the photon is completely transferred to the electron. If this energy is sufficient, the electrons are ejected out of the metal instantaneously .The minimum energy needed for the electron to come out of the metal surface is called work function. If the energy \[h\nu \] of the incident photon exceeds the work function \[\left( {{\phi _0}} \right)\] ,the electrons are emitted with a maximum kinetic energy.

\[ \Rightarrow h\nu = {f_0} + {K_{\max }}\]

\[ \Rightarrow {K_{\max }} = h\nu - {f_0}\]

-For a given metal and frequency of incident radiation the photoelectric current is directly proportional to the intensity of incident radiation.

-The photoelectric effect is an instantaneous process.

-The maximum kinetic energy of the emitted photoelectrons is independent of the intensity of the incident radiation but depends only on the frequency of the incident radiation.

**Complete step by step answer:**Given: Wavelength of the laser beam,

\[ \Rightarrow \lambda = 6600angstorm\]

$ \Rightarrow 6600 \times {10^{ - 10}}m$

Energy of each photon in the light beam is given by,

$E = h\nu $……………………(1)

Where, $h$ is the Planck’s constant

\[\upsilon \]is the frequency of a laser beam.

We know that, $c = \nu \lambda $

$ \Rightarrow \nu = \dfrac{c}{\lambda }$

Here $C$ is the speed of light.

Equation (1) can also be written as,

$ \Rightarrow E = h\dfrac{c}{\lambda }$

Now substitute the given values, we get

$ \Rightarrow E = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{6600 \times {{10}^{ - 10}}}}$

$ \Rightarrow E = 3 \times {10^{ - 19}}J$

Power or intensity of a laser beam,

$ \Rightarrow P = 6mW$

$ \Rightarrow 6 \times {10^{ - 3}}W$

Now, the number of photons falling per second on the target is given by,

$ \Rightarrow n = \dfrac{P}{E}$

Substitute the values we get

$ \Rightarrow n = \dfrac{{6 \times {{10}^{ - 3}}W}}{{3 \times {{10}^{ - 19}}J}}$

$\therefore n = 2 \times {10^{16}}$ Photons per second.

Hence, the number of photons passing along path of beam in per second is $2 \times {10^{16}}$

**Additional information:**According to Einstein’s photoelectric equation,

When a light is incident on a metal, the photons having energy \[h\nu \] collide with electrons at the surface of the metal. During these collisions, the energy of the photon is completely transferred to the electron. If this energy is sufficient, the electrons are ejected out of the metal instantaneously .The minimum energy needed for the electron to come out of the metal surface is called work function. If the energy \[h\nu \] of the incident photon exceeds the work function \[\left( {{\phi _0}} \right)\] ,the electrons are emitted with a maximum kinetic energy.

\[ \Rightarrow h\nu = {f_0} + {K_{\max }}\]

\[ \Rightarrow {K_{\max }} = h\nu - {f_0}\]

**Note:**-For a given metal and frequency of incident radiation the photoelectric current is directly proportional to the intensity of incident radiation.

-The photoelectric effect is an instantaneous process.

-The maximum kinetic energy of the emitted photoelectrons is independent of the intensity of the incident radiation but depends only on the frequency of the incident radiation.

Recently Updated Pages

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

The provincial president of the constituent assembly class 11 social science CBSE

Gersoppa waterfall is located in AGuyana BUganda C class 9 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

The hundru falls is in A Chota Nagpur Plateau B Calcutta class 8 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE