Q: Write Einstein's photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation.The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from \$'\lambda_{1}'\$ to \$'\lambda_{2}'\$. Derive the expressions for the threshold wavelength \$'\lambda_{0}'\$ and work function for the metal surface

Einstein's photoelectric equation states,\$h\nu=W_0+KE\$where \$\nu \$ is the frequency of incident light, \$W_0\$ is the work potential of the metal, and \$KE\$ is the maximum kinetc energy of released electron.The above equations explains the following results:1. If \$\nu \nu_o\$.2. The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectron depends only on the frequency of incident light.From given data,\$\dfrac{hc}{\lambda_1} = KE + W_o\$\$\dfrac{hc}{\lambda_2} = 2KE + W_o\$Eliminating KE, work function is given by:\$W_o = \dfrac{2hc}{\lambda_1} - \dfrac{hc}{\lambda_2}\$Threshold wavelength is found by:\$\dfrac{hc}{\lambda_o} = W_o\$Using above equations,\$\lambda_o = \dfrac{\lambda_1 \lambda_2}{2 \lambda_2 - \lambda_1}\$

Q: Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..

Work function of plate \$\phi = 1.9 \ eV\$Wavelength of incident photon \$\lambda = 5000 \ A^o = 500 \ nm\$Energy of the incident photon \$E_{incident} = \dfrac{1240}{\lambda (in \ nm)} \ eV\$\$\implies \ E_{incident} = \dfrac{1240}{500} = 2.48 \ eV\$Maximum kinetic energy of photon emitted \$K.E_{max} = E_{incident} - \phi\$\$\implies \ K.E_{max} = 2.48-1.9 = 0.58 \ eV\$

Q: When the light of frequency \$2v_0\$ (where \$v_0\$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is \$v_1\$. When the frequency of the incident radiation is increased to \$5v_0\$, the maximum velocity of electrons emitted from the same plate is \$v_2\$. The ratio of \$v_1\$ to \$v_2\$ is?

From photoelectric equation, \$h \nu = h \nu_o + \dfrac{1}{2} m v^2\$\$2h \nu_o = h \nu_o + \dfrac{1}{2} m v_1^2\$\$ \dfrac{1}{2} m v_1^2 = h \nu_o\$ ............... (1)Similarly, \$\dfrac{1}{2} m v_2^2 = 4 h \nu_o \$ .................(2)From (1) and (2), \$\dfrac{{v_1}^2}{{v_2}^2} = \dfrac{1}{4}\$\$\dfrac{v_1}{v_2} = \dfrac{1}{2}\$

Q: In the stydy of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:(i) Which one of the two metals has higher threshold frequency?(ii) Determine the work function of the metal which has greater value.(iii) Find the maximum kinetic energy of electron emitted by light of frequency \$8\times10^{14}Hz\$ for this metal.

(i) Threshold frequency of P is \$3\times10^{14}Hz.\$Threshold frequency of Q is \$6\times10^{14}Hz.\$Clearly Q has higher threshold frequency.(ii) Work function of metal Q,\$\psi_0=hv_0\$\$=(6.6\times10^{-34})\times6\times10^{14}J\$\$=\dfrac{39.6\times10^{-20}}{1.6\times10^{-19}}eV=2.5eV\$(iii) Maximum kinetic energy, \$K_{max}=hv-hv_0\$\$=h(v-v_0)\$\$=6.6\times10^{-34}(8\times10^{14}-6\times10^{14})\$\$=6.6\times10^{-34}\times2\times10^{14}J\$\$=\dfrac{6.6\times10^{-34}\times2\times10^{14}J}{1.6\times10^{-19}}eV\$\$\therefore K_{max}=0.83eV\$

Q: In a photoemissive cell with exciting wavelength \$\lambda \$, the fastest electron has speed \$v\$. If the exciting wavelength is changed by \$3\lambda /4\$, the speed of the fastest emitted electron will be :

From \$E={ W }_{ 0 }+\cfrac { 1 }{ 2 } m{ v }_{ max }^{ 2 }\Rightarrow { v }_{ max }=\sqrt { \cfrac { 2E }{ m } -\cfrac { 2{ W }_{ 0 } }{ m } } \$where \$E=\cfrac { hc }{ \lambda } \$If wavelength of incident light charges from \$\lambda\$ to \$\cfrac { 3\lambda }{ 4 } \$ (Decreases)Let energy of incident light charges from \$E\$ and speed of fastest electron changes from \$v\$ to \$v'\$ then\$\quad v'=\sqrt { \cfrac { 2E' }{ m } -\cfrac { 2{ W }_{ 0 } }{ m } } \$As \$E\propto \cfrac { 1 }{ \lambda } \Rightarrow E'\cfrac { 4 }{ 3 } E\$Hence \$v'=\sqrt { \cfrac { 2(\cfrac { 4 }{ 3 } )E }{ m } -\cfrac { 2{ W }_{ 0 } }{ m } } \$\$\Rightarrow v'={ \left( 4/3 \right) }^{ 1/2 }\sqrt { \cfrac { 2E }{ m } -\cfrac { 2{ W }_{ 0 } }{ m{ \left( 4/3 \right) }^{ 1/2 } } } \$So, \$v'>{ \left( 4/3 \right) }^{ 1/2 }v\$

Q: The slope of graph drawn between stopping potential and frequency of incident light for a given surface will be:-

\$eV_{o}= h V [V_{o}= Stooping\ potential, V= frequency ]\$\$\dfrac{V_{o}}{V}= \dfrac{h}{e} = \$ side

Question 1

Derive Einstein's photoelectric equation.

Accepted Answer

Einstein's photoelectric equation.According to Einstein, the emission of photoelectron is the result of the interaction between a single photon of the incident radiation and an electron in the metal.When a photon of energy $hv$ is incident on a metal surface, its energy is used up in two ways.(i) A part of the energy of the photon is used in extracting the electron from the surface of metal, since the electrons in the metal are bound to the nucleus. This energy $W$ spent in releasing the photoelectron is known as photoelectric work function of the metal. The work function of a photo metal is defined as the minimum amount of energy required to liberate an electron from the metal surface.(ii) The remaining energy of the photon is used to impart kinetic energy to the liberated electron. If $m$ is the mass of an electron and $v$ its velocity, thenEnergy of the incident photon $=$ Work function $+$ Kinetic energy of the electron$h\gamma = W + \dfrac {1}{2}mv^{2} .... (1)$If the electron does not lose energy by internal collisions, as it escapes from the metal, the entire energy $(h - W)$ will be exhibited as the kinetic energy of the electron Thus, $(h - W)$ represents the maximum kinetic energy of the ejected photoelectron. If Vman is the maximum velocity with which the photoelectron can be ejected, then$hv = W + \dfrac {1}{2} mv^{2}_{max} .... (2)$This equation is known as Einstein's photoelectric equation.When the frequency $(\gamma)$ of the incident radiation is equal to the threshold frequency $(\gamma_{0})$ of the metal surface, kinetic energy of the electron is zero. Then equation $(2)$ becomes,$h\gamma_{0} = W .... (3)$Substituting the value of $W$ in equation $(2)$ we get,$h\gamma - h\gamma_{0} = mv_{max}^{2}$ (or) $h(\gamma - \gamma_{0}) = mv_{max}^{2}$This is another form of Einstein's photoelectric equation.

Question 2

Write Einstein's photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation.
The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from $^{'} λ_{1}^{'}$ to $^{'} λ_{2}^{'}$ . Derive the expressions for the threshold wavelength $^{'} λ_{0}^{'}$ and work function for the metal surface

Accepted Answer

Einstein's photoelectric equation states,$h\nu=W_0+KE$where $\nu $ is the frequency of incident light, $W_0$ is the work potential of the metal, and $KE$ is the maximum kinetc energy of released electron.The above equations explains the following results:1.  If $\nu < \nu_o$, then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if $\nu > \nu_o$.2. The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectron depends only on the frequency of incident light.From given data,$\dfrac{hc}{\lambda_1} = KE + W_o$$\dfrac{hc}{\lambda_2} = 2KE + W_o$Eliminating KE, work function is given by:$W_o = \dfrac{2hc}{\lambda_1} - \dfrac{hc}{\lambda_2}$Threshold wavelength is found by:$\dfrac{hc}{\lambda_o} = W_o$Using above equations,$\lambda_o = \dfrac{\lambda_1 \lambda_2}{2 \lambda_2 - \lambda_1}$

Question 3

Draw a plot showing the variation of photoelectric current with collector plate potential foe two different frequencies, $v_1 > v_2$, of incident radiation having the same intensity. In which case will the stopping potential be higher ? Justify your answer.

Accepted Answer

The variation of photoelectric current with collector potential for two different frequencies, $\nu_1>\nu_2$ is shown in figure.The photoelectric equation in terms of stopping potential $V_0$ is given by , $eV_0=h\nu-h\nu_o$So, $V_0 \propto \nu$  Thus, Stopping potential will be higher for $\nu_1$.

Question 4

Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..

Accepted Answer

Work function of plate  $\phi = 1.9 \ eV$Wavelength of incident photon  $\lambda = 5000 \ A^o = 500 \ nm$Energy of the incident photon   $E_{incident} = \dfrac{1240}{\lambda (in \ nm)} \ eV$$\implies \ E_{incident} = \dfrac{1240}{500} = 2.48 \ eV$Maximum kinetic energy of photon emitted   $K.E_{max} = E_{incident} - \phi$$\implies \ K.E_{max} = 2.48-1.9 = 0.58 \ eV$

Question 5

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$, the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is?

Accepted Answer

From photoelectric equation, $h \nu = h \nu_o + \dfrac{1}{2} m v^2$$2h \nu_o = h \nu_o + \dfrac{1}{2} m v_1^2$$ \dfrac{1}{2} m v_1^2 = h \nu_o$ ............... (1)Similarly, $\dfrac{1}{2} m v_2^2 = 4 h \nu_o $ .................(2)From (1) and (2), $\dfrac{{v_1}^2}{{v_2}^2} = \dfrac{1}{4}$$\dfrac{v_1}{v_2} = \dfrac{1}{2}$

Question 6

Which of the following phenomenon works on the particle nature of light?

Accepted Answer

(a) Interference is a phenomenon in which two waves of same frequency superpose to give resultant intensity different from sum of their separate intensity. So, it cannot exhibit particle's nature of light.(b) Diffraction is a phenomenon in which light bends at sharp ends of an obstacle or a hole. So, it also cannot exhibit particle's nature of light.(c) Polarisation of light is a property owing to which a light ray after emerging through a crystal (a special kind like tourmaline) have vibrations in a plane perpendicular to its direction of propagation. So, it also can not explain particle's nature of light.(d) Photoelectric effect states that light travels in the form of bundles or packets of energy, called photons. This effect is explained on the basis of quantum nature of light. So, it clearly explains the particles's nature of lighthence (d) is correct.

Question 7

In the stydy of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:
(i) Which one of the two metals has higher threshold frequency?
(ii) Determine the work function of the metal which has greater value.
(iii) Find the maximum kinetic energy of electron emitted by light of frequency $8 \times 1 0^{14} H z$ for this metal.

Accepted Answer

(i) Threshold frequency of P is $3\times10^{14}Hz.$Threshold frequency of Q is $6\times10^{14}Hz.$Clearly Q has higher threshold frequency.(ii) Work function of metal Q,$\psi_0=hv_0$$=(6.6\times10^{-34})\times6\times10^{14}J$$=\dfrac{39.6\times10^{-20}}{1.6\times10^{-19}}eV=2.5eV$(iii) Maximum kinetic energy, $K_{max}=hv-hv_0$$=h(v-v_0)$$=6.6\times10^{-34}(8\times10^{14}-6\times10^{14})$$=6.6\times10^{-34}\times2\times10^{14}J$$=\dfrac{6.6\times10^{-34}\times2\times10^{14}J}{1.6\times10^{-19}}eV$$\therefore K_{max}=0.83eV$

Question 8

In a photoemissive cell with exciting wavelength $\lambda $, the fastest electron has speed $v$. If the exciting wavelength is changed by $3\lambda /4$, the speed of the fastest emitted electron will be :

Accepted Answer

From $E={ W }_{ 0 }+\cfrac { 1 }{ 2 } m{ v }_{ max }^{ 2 }\Rightarrow { v }_{ max }=\sqrt { \cfrac { 2E }{ m } -\cfrac { 2{ W }_{ 0 } }{ m }  } $where $E=\cfrac { hc }{ \lambda  } $If wavelength of incident light charges from $\lambda$ to $\cfrac { 3\lambda  }{ 4 } $ (Decreases)Let energy of incident light charges from $E$ and speed of fastest electron changes from $v$ to $v'$ then$\quad v'=\sqrt { \cfrac { 2E' }{ m } -\cfrac { 2{ W }_{ 0 } }{ m }  } $As $E\propto \cfrac { 1 }{ \lambda  } \Rightarrow E'\cfrac { 4 }{ 3 } E$Hence $v'=\sqrt { \cfrac { 2(\cfrac { 4 }{ 3 } )E }{ m } -\cfrac { 2{ W }_{ 0 } }{ m }  } $$\Rightarrow v'={ \left( 4/3 \right)  }^{ 1/2 }\sqrt { \cfrac { 2E }{ m } -\cfrac { 2{ W }_{ 0 } }{ m{ \left( 4/3 \right)  }^{ 1/2 } }  } $So, $v'>{ \left( 4/3 \right)  }^{ 1/2 }v$

Question 9

The graph between the stopping potential $V_0$ and frequency $v$ for two different metal plates P and Q are shown in the figure. Which of the metals has greater threshold wavelength and work function?

Accepted Answer

The threshold frequency of metal plate P is lesser than that of Q.Since $\lambda\propto \dfrac{1}{\nu}$, the threshold wavelength of metal plate P is greater than that of Q.The work function of a metal is given as $W_0=h\nu_0$Since the threshold frequency of metal Q is greater, the work function it is also greater than that of P.

Question 10

The slope of graph drawn between stopping potential and frequency of incident light for a given surface will be:-

Accepted Answer

$eV_{o}= h V [V_{o}= Stooping\ potential, V= frequency  ]$$\dfrac{V_{o}}{V}= \dfrac{h}{e} = $ side

Revise with Concepts

Photo Electric Effect of Light and Wave Theory of Light

Einstein's Photoelectric Equation

Particle Nature of Light

Quick Summary With Stories

Photoelectric Effect - III

Einstein’s Photoelectric Equation.

The Light Is Having Particle Nature As Well As Wave Nature(dual nature of light )

Particle Nature of Light - Problem L1

Derive Einstein's photoelectric equation.

Draw a plot showing the variation of photoelectric current with collector plate potential foe two different frequencies, $v_{1} > v_{2}$ , of incident radiation having the same intensity. In which case will the stopping potential be higher ? Justify your answer.

Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..

Which of the following phenomenon works on the particle nature of light?

In a photoemissive cell with exciting wavelength $λ$ , the fastest electron has speed $v$ . If the exciting wavelength is changed by $3 λ / 4$ , the speed of the fastest emitted electron will be :

The graph between the stopping potential $V_{0}$ and frequency $v$ for two different metal plates P and Q are shown in the figure. Which of the metals has greater threshold wavelength and work function?

The slope of graph drawn between stopping potential and frequency of incident light for a given surface will be:-

Revise with Concepts

Photo Electric Effect of Light and Wave Theory of Light

Einstein's Photoelectric Equation

Particle Nature of Light

Quick Summary With Stories

Photoelectric Effect - III

Einstein’s Photoelectric Equation.

The Light Is Having Particle Nature As Well As Wave Nature(dual nature of light )

Particle Nature of Light - Problem L1

Derive Einstein's photoelectric equation.

Draw a plot showing the variation of photoelectric current with collector plate potential foe two different frequencies, v1​>v2​, of incident radiation having the same intensity. In which case will the stopping potential be higher ? Justify your answer.

Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..

Which of the following phenomenon works on the particle nature of light?

In a photoemissive cell with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed by 3λ/4, the speed of the fastest emitted electron will be :

The graph between the stopping potential V0​ and frequency v for two different metal plates P and Q are shown in the figure. Which of the metals has greater threshold wavelength and work function?

The slope of graph drawn between stopping potential and frequency of incident light for a given surface will be:-

Draw a plot showing the variation of photoelectric current with collector plate potential foe two different frequencies, $v_{1} > v_{2}$ , of incident radiation having the same intensity. In which case will the stopping potential be higher ? Justify your answer.

In a photoemissive cell with exciting wavelength $λ$ , the fastest electron has speed $v$ . If the exciting wavelength is changed by $3 λ / 4$ , the speed of the fastest emitted electron will be :

The graph between the stopping potential $V_{0}$ and frequency $v$ for two different metal plates P and Q are shown in the figure. Which of the metals has greater threshold wavelength and work function?