Q: Light of two different frequencies whose photons have energies 1eV and 2.5eV successively illuminate a metal of work function 0.5 eV.The ratio of the maximum speeds of the emitted electrons will he

For maximum speed of the photo electrons \$\dfrac{1}{2}mv^2 = E_{photon} - \phi\$where \$\phi = 0.5\$ eV is the work function of the metal.Energy of photon \$E_1 = 1eV\$\$\therefore\$ \$\dfrac{1}{2}mv_1^2 =1- 0.5\$We get \$\dfrac{1}{2}mv_1^2 =0.5\$ eV ....(1)Energy of photon \$E_2 = 2.5eV\$\$\therefore\$ \$\dfrac{1}{2}mv_2^2 =2.5- 0.5\$We get \$\dfrac{1}{2}mv_2^2 =2\$ eV ....(2)Ratio of maximum speeds \$\dfrac{v_1^2}{v_2^2} = \dfrac{0.5}{2} = \dfrac{1}{4}\$We get \$\dfrac{v_1}{v_2} = \dfrac{1}{2}\$

Q: The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work-function? Justify your answer.

\$eV_0=hv-hv_0\$here, e is the charge of electron, h=planck's constant, \$v_0\$ is the threshold frequency and \$v\$ is the frequency of the incident lighton comparing with the \$Y=mX+C \$ we can observe that more the intercept on the Y-axis or \$V_0\$ more is the frequency(\$v_0\$).Hence from the graph, the threshold frequency of Metal A is greater than the Metal B, therefore the work function of Metal A is more than Metal B.

Q: A photoelectric surface is illuminated successively by monochromatic light of wavelength \$\lambda\$ and \$\dfrac{\lambda }{2}\$. If the maximum kinetic energy of the emitted photoelectrons in the second case is \$3\$ times that in the first case, the work function of the surface of the material is :(\$h =\$ Planck's constant, \$c = \$speed of light)

In first case, \$E_0=\dfrac{hc}{\lambda}-\phi ......(1)\$and in 2nd case, \$3E_0=\dfrac{hc}{\lambda/2}-\phi .....(2)\$\$(2)/(1) \Rightarrow \dfrac{3}{1}=\dfrac{\dfrac{2hc}{\lambda}-\phi}{\dfrac{hc}{\lambda}-\phi}\$or \$\dfrac{3hc}{\lambda}-3\phi=\dfrac{2hc}{\lambda}-\phi\$or \$\phi=\dfrac{hc}{2\lambda}\$

Q: A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is

Energy of incident light \$E\left( eV \right) =\dfrac { 12375 }{ 3320 } =3.72eV\$We know that, \$E={ W }_{ 0 }+{ eV }_{ 0 }\$\${ V }_{ 0 }=\dfrac { \left( E-{ W }_{ 0 } \right) }{ e } =\dfrac { 3.72eV-1.07eV }{ e } \$\${ V }_{ 0 }=2.65\$ Volt

Q: The maximum kinetic energy of the photoelectrons emitted is doubled when the wavelength of light incident on the photosensitive surface changes from \$\lambda_1\$ to \$\lambda_2\$. Deduce expressions for the threshold wavelength and work function for the metal surface in terms of \$\lambda_1\$ and \$\lambda_2\$.

Given that Initial kinetic energy of photoelectrons \$= K_1\$final kinetic energy of photoelectrons \$= 2 K_1\$wavelength of light changes from \$\lambda_1\$ to \$\lambda_2\$Let the threshold frequency is \$\lambda_0\$ and work function is \$\omega_0\$Now, we know\$\dfrac{h c}{\lambda} = \omega_0 + KE\$\$\dfrac{h c}{\lambda_1} = \omega_0 + K_1\$ ___(I)\$\dfrac{h c}{\lambda_2} = \omega_0 + K_2\$ __(II)\$K_2 = 2 K_1\$\$\dfrac{hc}{\lambda_2} = \omega_0 + 2K_1\$ ___(III)\$\dfrac{2 hc}{\lambda_1} = 2 \omega_0 + 2 K_1\$ (eq (I) \$\times \$ 2)\$\dfrac{2 hc}{\lambda_1} - \dfrac{h c}{\lambda_2} = \omega_0\$ \$\Rightarrow \omega_0 = hc \left(\dfrac{2 \lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right)\$we knowwork function \$\omega_0 = \dfrac{hc}{\lambda_0}\$\$\dfrac{hc}{\lambda_0} hc \left(\dfrac{2 \lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right)\$\$\dfrac{1}{\lambda_0} = \dfrac{2 \lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\$\$\lambda_0 = \dfrac{\lambda_1 \lambda_2}{2 \lambda_2 - \lambda_1}\$

Q: Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of \$3.3 V\$. If photo electric emission begins in this metal at a frequency of \$8 \times 10^{14} Hz\$, calculate the work function (in eV) for this metal.

Given that;Stoping potential, \$v_s=3.3\$vMinimum frequency needed to eject electrons from metal\$=\$ frequency for work function \$=\nu_{Th}=\$threshold frequency\$\nu_{Th}=8\times 10^{14}Hz\$\$\therefore \phi=\$ work function \$=h\nu_{Th}=6.62\times 10^{-34}\times 8\times 10^{14}\$\$\phi =5.296\times 10^{-19}J\$\$\because 1.6\times 10^{-19}J=1eV\$\$\phi =\dfrac{5.296\times 10^{-19}}{1.6\times 10^{-19}}=3.31eV\$Work function for the metal \$=\phi =3.31\$eVUse Einstein equation:-Incident energy, \$E=\phi +ev_s\$\$\Rightarrow E=(3.31+3.3)eV\$\$E=6.61ev=(6.61\times 10^{-19}\times 1.6)J\$\$E=h\nu=6.61\times 1.6\times 10^{-19}\$\$\Rightarrow \nu=\dfrac{6.61\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}=1.6\times 10^{15}Hz\$Thus, frequency of light \$=\nu=1.6\times 10^{15}Hz\$.

Question 1

Light of two different frequencies whose photons have energies 1eV and 2.5eV successively illuminate a metal of work function 0.5 eV.The ratio of the maximum speeds of the emitted electrons will he

Accepted Answer

For maximum speed of the photo electrons  $\dfrac{1}{2}mv^2  = E_{photon} - \phi$where $\phi = 0.5$ eV is the work function of the metal.Energy of photon $E_1 = 1eV$$\therefore$  $\dfrac{1}{2}mv_1^2  =1- 0.5$We get  $\dfrac{1}{2}mv_1^2  =0.5$ eV        ....(1)Energy of photon $E_2 = 2.5eV$$\therefore$  $\dfrac{1}{2}mv_2^2  =2.5- 0.5$We get  $\dfrac{1}{2}mv_2^2  =2$ eV        ....(2)Ratio of maximum speeds  $\dfrac{v_1^2}{v_2^2} = \dfrac{0.5}{2} = \dfrac{1}{4}$We get   $\dfrac{v_1}{v_2} =  \dfrac{1}{2}$

Question 2

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 3

The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work-function? Justify your answer.

Accepted Answer

$eV_0=hv-hv_0$here, e is the charge of electron, h=planck's constant, $v_0$ is the threshold frequency and $v$ is the frequency of the incident lighton comparing with the $Y=mX+C $  we can observe that more the intercept on the Y-axis or $V_0$ more is the frequency($v_0$).Hence from the graph, the threshold frequency of Metal A is greater than the Metal B, therefore the work function of Metal A is more than Metal B.

Question 4

A photoelectric surface is illuminated successively by monochromatic light of wavelength $λ$ and $\frac{λ}{2}$ . If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case, the work function of the surface of the material is :
( $h =$ Planck's constant, $c =$ speed of light)

Accepted Answer

In first case, $E_0=\dfrac{hc}{\lambda}-\phi   ......(1)$and in 2nd case,  $3E_0=\dfrac{hc}{\lambda/2}-\phi    .....(2)$$(2)/(1) \Rightarrow \dfrac{3}{1}=\dfrac{\dfrac{2hc}{\lambda}-\phi}{\dfrac{hc}{\lambda}-\phi}$or $\dfrac{3hc}{\lambda}-3\phi=\dfrac{2hc}{\lambda}-\phi$or $\phi=\dfrac{hc}{2\lambda}$

Question 5

What is matter waves? Write any two characteristics of it.

Accepted Answer

Matter waves: According to De-Broglie, a wave is associated with each moving particle which is called matter waves. This wave has wave length &#955;=hp&#955;=hp. Here h is Planck's constant and p is the moment of the moving particle.Characteristics of Matter waves:(i) Lighter the particle, greater is the De-Broglie wave length.(ii) The faster the particle moves, the smaller is its De-Broglie wave length.(iii) The De-Broglie wave length of a particle in independent of the charge or nature of the particle.(iv) The matter waves are not electromagnetic in nature. The electromagnetic waves are produced only by charged particle.

Question 6

When the energy of the incident radiation is increased by $20$%, the kinetic energy of the photoelectrons emitted from a metal surface increased from emitted $0.5 $eV to $0.8$ eV. The work function of the metal is

Accepted Answer

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Question 7

A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is

Accepted Answer

Energy of incident light $E\left( eV \right) =\dfrac { 12375 }{ 3320 } =3.72eV$We know that, $E={ W }_{ 0 }+{ eV }_{ 0 }$${ V }_{ 0 }=\dfrac { \left( E-{ W }_{ 0 } \right)  }{ e } =\dfrac { 3.72eV-1.07eV }{ e } $${ V }_{ 0 }=2.65$ Volt

Question 8

The maximum kinetic energy of the photoelectrons emitted is doubled when the wavelength of light incident on the photosensitive surface changes from $\lambda_1$ to $\lambda_2$. Deduce expressions for the threshold wavelength and work function for the metal surface in terms of $\lambda_1$ and $\lambda_2$.

Accepted Answer

Given that Initial kinetic energy of photoelectrons $= K_1$final kinetic energy of photoelectrons $= 2 K_1$wavelength of light changes from $\lambda_1$ to $\lambda_2$Let the threshold frequency is $\lambda_0$ and work function is $\omega_0$Now, we know$\dfrac{h c}{\lambda} = \omega_0 + KE$$\dfrac{h c}{\lambda_1} = \omega_0 + K_1$ ___(I)$\dfrac{h c}{\lambda_2} = \omega_0 + K_2$ __(II)$K_2 = 2 K_1$$\dfrac{hc}{\lambda_2} = \omega_0 + 2K_1$ ___(III)$\dfrac{2 hc}{\lambda_1} = 2 \omega_0 + 2 K_1$    (eq (I) $\times $ 2)$\dfrac{2 hc}{\lambda_1} - \dfrac{h c}{\lambda_2} = \omega_0$ $\Rightarrow \omega_0 = hc \left(\dfrac{2 \lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right)$we knowwork function $\omega_0 = \dfrac{hc}{\lambda_0}$$\dfrac{hc}{\lambda_0} hc \left(\dfrac{2 \lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right)$$\dfrac{1}{\lambda_0} = \dfrac{2 \lambda_2 - \lambda_1}{\lambda_1 \lambda_2}$$\lambda_0 = \dfrac{\lambda_1 \lambda_2}{2 \lambda_2 - \lambda_1}$

Question 9

Find the frequency of light  which ejects electrons from a metal surface, fully stopped by a retarding potential of $3.3 V$. If photo electric emission begins in this metal at a frequency of $8 \times 10^{14} Hz$, calculate the work function (in eV) for this metal.

Accepted Answer

Given that;Stoping potential, $v_s=3.3$vMinimum frequency needed to eject electrons from metal$=$ frequency for work function $=\nu_{Th}=$threshold frequency$\nu_{Th}=8\times 10^{14}Hz$$\therefore \phi=$ work function $=h\nu_{Th}=6.62\times 10^{-34}\times 8\times 10^{14}$$\phi =5.296\times 10^{-19}J$$\because 1.6\times 10^{-19}J=1eV$$\phi =\dfrac{5.296\times 10^{-19}}{1.6\times 10^{-19}}=3.31eV$Work function for the metal $=\phi =3.31$eVUse Einstein equation:-Incident energy, $E=\phi +ev_s$$\Rightarrow E=(3.31+3.3)eV$$E=6.61ev=(6.61\times 10^{-19}\times 1.6)J$$E=h\nu=6.61\times 1.6\times 10^{-19}$$\Rightarrow \nu=\dfrac{6.61\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}=1.6\times 10^{15}Hz$Thus, frequency of light $=\nu=1.6\times 10^{15}Hz$.

Question 10

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

Light of two different frequencies whose photons have energies 1eV and 2.5eV successively illuminate a metal of work function 0.5 eV.The ratio of the maximum speeds of the emitted electrons will he

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

The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work-function? Justify your answer.

What is matter waves? Write any two characteristics of it.

When the energy of the incident radiation is increased by $20$ %, the kinetic energy of the photoelectrons emitted from a metal surface increased from emitted $0.5$ eV to $0.8$ eV. The work function of the metal is

A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is

The maximum kinetic energy of the photoelectrons emitted is doubled when the wavelength of light incident on the photosensitive surface changes from $λ_{1}$ to $λ_{2}$ . Deduce expressions for the threshold wavelength and work function for the metal surface in terms of $λ_{1}$ and $λ_{2}$ .

Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of $3.3 V$ . If photo electric emission begins in this metal at a frequency of $8 \times 1 0^{14} H z$ , calculate the work function (in eV) for this metal.

Obtains Bohr's quantisation condition for angular momentum of electron orbiting in $n^{t h}$ hydrogen atom on the basis of the wave picture of an electron using de Broglie hypothesis.

Light of two different frequencies whose photons have energies 1eV and 2.5eV successively illuminate a metal of work function 0.5 eV.The ratio of the maximum speeds of the emitted electrons will he

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

The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work-function? Justify your answer.

What is matter waves? Write any two characteristics of it.

When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from emitted 0.5eV to 0.8 eV. The work function of the metal is

A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is

The maximum kinetic energy of the photoelectrons emitted is doubled when the wavelength of light incident on the photosensitive surface changes from λ1​ to λ2​. Deduce expressions for the threshold wavelength and work function for the metal surface in terms of λ1​ and λ2​.

Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of 3.3V. If photo electric emission begins in this metal at a frequency of 8×1014Hz, calculate the work function (in eV) for this metal.

Obtains Bohr's quantisation condition for angular momentum of electron orbiting in nth hydrogen atom on the basis of the wave picture of an electron using de Broglie hypothesis.

When the energy of the incident radiation is increased by $20$ %, the kinetic energy of the photoelectrons emitted from a metal surface increased from emitted $0.5$ eV to $0.8$ eV. The work function of the metal is

The maximum kinetic energy of the photoelectrons emitted is doubled when the wavelength of light incident on the photosensitive surface changes from $λ_{1}$ to $λ_{2}$ . Deduce expressions for the threshold wavelength and work function for the metal surface in terms of $λ_{1}$ and $λ_{2}$ .

Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of $3.3 V$ . If photo electric emission begins in this metal at a frequency of $8 \times 1 0^{14} H z$ , calculate the work function (in eV) for this metal.

Obtains Bohr's quantisation condition for angular momentum of electron orbiting in $n^{t h}$ hydrogen atom on the basis of the wave picture of an electron using de Broglie hypothesis.