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: 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: 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: 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\$

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

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 3

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 4

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 5

Obtains Bohr's quantisation condition for angular momentum of electron orbiting in $n^{th}$ hydrogen atom on the basis of the wave picture of an electron using de Broglie hypothesis.

Accepted Answer

According to De-brogile's wavelength associated with an electron ;$\lambda =\dfrac{h}{mv}$......1and the standing wave condition that circumference $=$ whole Number of wave length If 'n' is the nth orbit$\therefore 2\pi r=n\lambda n$....2$\therefore $ since angular momentum $L=mvr$ from eqn (1)$L=\dfrac{hr}{\lambda}$$=\dfrac{hr}{\left(\dfrac{2\pi r}{n}\right)}$ from 1 and 2$\therefore L=\dfrac{nh}{2\pi}$ Bohr's Quantization condition

Question 6

Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions  $W _ { 1 }$  and  $W _ { 2 } \left( W _ { 1 } > W _ { 2 } \right) .$  On what factors does the (i) slope and (ii) intercept of the line depend?

Accepted Answer

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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

Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies $V_{A} > V_{B}$
(i) In which case is the stopping potential more and why?
(ii) Does the slope of the graph depend on the nature of the material used? Explain

Accepted Answer

(i) From the graph for the same value of 'v', stopping potential is more for material 'B' From Einstein's photoelectric equation$eV_0=hv-hv_0$$V_0 =\dfrac{h}{e}-\dfrac{h}{e}_0=\dfrac{h}{e}(v-v_0)$Therefore$ V_0$ is higher for lower value of $V_0$ii. No, as slope is given by $\dfrac{h}{e}$ which is a universal constant.

Question 9

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 10

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, $v'>{ \left( 4/3 \right)  }^{ 1/2 }v$

Einstein's Photoelectric Equation

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