Q: The photoelectric work function for a metal surface is 4.125eV. The cut-off wavelength for this surface is

Since work function for a metal surface is \$W=\dfrac{hc}{\lambda_0}\$where \$\lambda_0\$ is thershold wavelength or cut-off wavelength for a metral surface.here \$W=4.125 e V=4.125\times 1.6\times 10^{-19}\, Joule\$so \$\lambda _0=\dfrac{6.6\times 10^{-34}\times 3\times 10^8 }{4.125 \times 1.6 \times 10^{-19}}=3000 \mathring {A } \$

Q: Photons with energy \$5\ eV\$ are incident on a cathode \$C\$ in a photoelectric cell. The maximum energy of emitted photoelectrons is \$2\ eV\$. When photons of energy \$6\ eV\$ are incident on \$C\$, no photoelectrons will reach the anode \$A\$, if the stopping potential of \$A\$ relative to \$C\$ is

When \$5eV\$ is incident the kinetic energy is \$2eV\$ it simply means the work function is \$W=5eV-2eV=3eV\$Similarly When \$6eV\$ is incident the kinetic energy should be \$6eV-W=6eV-3eV=3eV\$ it simply means to stop them we need a \$negative \$ potential at anode equal to \$\dfrac{3eV}{e}=3V\$So answer is \$-3V\$ i.e. option B is correct.

Q: When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are respectively 0.6 V and 18 mA. If the same source is placed 0.6 m away from the photocell, then :

Intensity of photons \$\uparrow\$, no. of electron \$\uparrow\$The cut off voltage does not depend on distance between cell and source. But intensity is inversely proportional to square of it's distance. \$\therefore \$ at \$d = 0.6 m,\$Cut off voltage \$ = 0.6 V\$but \$I_s = 18\times (\dfrac{0.2}{0.6})^2\$\$= {2mA}\$So, the answers are option (C) & (D).

Q: The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200nm falls on it, must be

Work function of metal surface \$\phi = 5.01 \ eV\$Wavelength of incident photon \$\lambda = 200 \ nm\$Energy of incident photon \$E = \dfrac{1240}{\lambda \ (in nm)} \ eV = \dfrac{1240}{200} = 6.21 \ eV\$Stopping potential \$eV_o = E-\phi\$\$\therefore \ eV_o = 6.21 - 5.01 =1. 2eV\$\$\implies \ V_o = 1.2 \ V\$

Q: The threshold frequency for a photosensitive metal is \$3.3\times{10}^{4}\$Hz. If light of frequency \$8.2\times{10}^{14}\$Hz is incident on this metal. The cut-off voltage for the photoelectric emission is nearly

Given : \$\nu_{th} = 3.3\times 10^{14} \ Hz\$ \$\nu = 8.2\times 10^{14} \ Hz\$Cut-off voltage \$eV_o = h\nu - h\nu_{th}\$Or \$V_o = \dfrac{h}{e}(\nu - \nu_{th})\$\$\implies \ V_o = \dfrac{6.6\times 10^{-34}}{1.6\times 10^{-19}}(8.2-3.3)\times 10^{14} = 2 \ V\$

Q: In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :

Let \$V_0\$ be the stopping potentialIn a photoelectric effect, the maximum kinetic energy of electron , \$K_{max}=eV_0\$so, \$V_0=\dfrac{K_{max}}{e}=\dfrac{0.5eV}{e}=0.5 V\$ (as eV=1 electronic charge times 1 Volt)

Question 1

The photoelectric work function for a metal surface is 4.125eV. The cut-off wavelength for this surface is

Accepted Answer

Since work function for a metal surface is $W=\dfrac{hc}{\lambda_0}$where $\lambda_0$ is thershold wavelength or cut-off wavelength for a metral surface.here $W=4.125 e V=4.125\times 1.6\times 10^{-19}\, Joule$so $\lambda _0=\dfrac{6.6\times 10^{-34}\times 3\times 10^8 }{4.125 \times 1.6 \times 10^{-19}}=3000 \mathring {A } $

Question 2

A $12.5\ eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Accepted Answer

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is $12.5$ eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is $&#8722;13.6$ eV.  When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes $&#8722;13.6 + 12.5$ eV i.e., $&#8722;1.1$ eV.  Orbital energy is related to orbit level $(n)$ as:$E=\dfrac{-13.6}{{(n)}^{2}}$eVFor $n=3, E=\dfrac{-13.6}{9}=-1.5$eVThis energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from $n = 1$ to $n = 3$ level. During its de-excitation, the electrons can jump from $n = 3$ to $n = 1$ directly, which forms a line of the Lyman series of the hydrogen spectrum.  We have the relation for wave number for Lyman series as:$\dfrac{1}{\lambda}={R}_{y}\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{n}^{2}}\right)$Where, ${R}_{y}=$Rydberg constant $= 1.097 \times {10}^{7} {m}^{&#8722;1}$$\lambda=$ Wavelength of radiation emitted by the transition of the electron For $n = 3$, we can obtain $\lambda$ as: $\dfrac{1}{\lambda}=1.097\times{10}^{7}\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{3}^{2}}\right)=1.097\times{10}^{7}\times\dfrac{8}{9}$$\Rightarrow \lambda=\dfrac{9}{8\times 1.097\times{10}^{7}}=102.55$nmIf the electron jumps from $n = 2$ to $n = 1,$ then the wavelength of the radiation is given as:$\dfrac{1}{\lambda}=1.097\times{10}^{7}\left(\dfrac{1}{{2}^{1}}-\dfrac{1}{{2}^{2}}\right)=1.097\times{10}^{7}\times\dfrac{3}{4}$$\Rightarrow \lambda=\dfrac{4}{3\times 1.097\times{10}^{7}}=121.54$nmIf the electron jumps from $n =3$ to $n = 2$ then the wavelength of the radiation is given as:$\dfrac{1}{\lambda}=1.097\times{10}^{7}\left(\dfrac{1}{{2}^{1}}-\dfrac{1}{{3}^{2}}\right)=1.097\times{10}^{7}\times\dfrac{5}{46}$$\Rightarrow \lambda=\dfrac{36}{5\times 1.097\times{10}^{7}}=653.33$nmThis radiation corresponds to the Balmer series of the hydrogen spectrum.  Hence, in Lyman series, two wavelengths i.e., $102.5$ nm and $121.5$nm are emitted. And in the Balmer series, one wavelength i.e., $656.33$ nm is emitted.

Question 3

Photons with energy $5\ eV$ are incident on a cathode $C$ in a photoelectric cell. The maximum energy of emitted photoelectrons is $2\ eV$. When photons of energy $6\ eV$ are incident on $C$, no photoelectrons will reach the anode $A$, if the stopping potential of $A$ relative to $C$ is

Accepted Answer

When $5eV$ is  incident the kinetic energy is $2eV$ it simply means the work function is $W=5eV-2eV=3eV$Similarly When $6eV$ is incident the kinetic energy should be   $6eV-W=6eV-3eV=3eV$ it simply means to stop them we need a $negative $ potential at anode equal to $\dfrac{3eV}{e}=3V$So answer is $-3V$ i.e. option B is correct.

Question 4

Define stopping potential (cut-off potential) with reference to photoelectric effect.

Accepted Answer

Stopping potential is defined as the potential required to stop ejection of electron from a metal surface when incident beam of energy greater than the work potential of metal is directed on it.

Question 5

When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are respectively 0.6 V and 18 mA. If the same source is placed 0.6 m away from the photocell, then :

Accepted Answer

Intensity of photons $\uparrow$, no. of electron $\uparrow$The cut off voltage does not depend on distance between cell and source. But intensity is inversely proportional to square of it's distance. $\therefore $ at $d = 0.6 m,$Cut off voltage $ = 0.6 V$but $I_s = 18\times (\dfrac{0.2}{0.6})^2$$= {2mA}$So, the answers are option (C) & (D).

Question 6

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200nm falls on it, must be

Accepted Answer

Work function of metal surface  $\phi = 5.01 \ eV$Wavelength of incident photon  $\lambda = 200 \ nm$Energy of incident photon  $E = \dfrac{1240}{\lambda \ (in nm)} \ eV = \dfrac{1240}{200} = 6.21 \ eV$Stopping potential  $eV_o = E-\phi$$\therefore \ eV_o = 6.21 - 5.01 =1. 2eV$$\implies \ V_o = 1.2 \ V$

Question 7

In a photoelectric cell, the wavelength of incident light is changed from $4000\overset {o}{A}$ to $3000 \overset {o}{A}$. The change in stopping potential will be :

Accepted Answer

The Einstein's equation for photoelectric effect is given by $eV_0= \dfrac{hc}{\lambda}-W$  where $V_0=$ stopping potential, $h=$ Planck's constant, $c=$ velocity of light, $\lambda=$wavelength of incident light and $W=$ work function of metal.Here, $eV_{01}=\dfrac{hc}{\lambda_1}-W ....(1)$ and $eV_{02}=\dfrac{hc}{\lambda_2}-W ....(2)$From (1) and (2),  $eV_{02}=\dfrac{hc}{\lambda_2}+eV_{01}-\dfrac{hc}{\lambda_1}$$eV_{02}-eV_{01}=\dfrac{hc}{\lambda_2}-\dfrac{hc}{\lambda_1}$$V_{02}-V_{01}=(hc/e)[1/\lambda_2-1/\lambda_1]=\dfrac{6.62\times 10^{-34}\times 3\times 10}{1.6\times 10^{-19}}[\dfrac{1}{3000\times 10^{-10}}-\dfrac{1}{4000\times 10^{-10}}]=1.04 V$

Question 8

In photoelectric effect, stopping potential depends upon

Accepted Answer

The stopping potenial depends on frequency of incident light and nature &#10;of the emitter material.for a given frequency of incident light, it is &#10;independent of its intensity.

Question 9

The threshold frequency for a photosensitive metal is $3.3\times{10}^{4}$Hz. If light of frequency $8.2\times{10}^{14}$Hz is incident on this metal. The cut-off voltage for the photoelectric emission is nearly

Accepted Answer

Given :   $\nu_{th} = 3.3\times 10^{14} \ Hz$         $\nu = 8.2\times 10^{14} \ Hz$Cut-off voltage  $eV_o = h\nu  - h\nu_{th}$Or   $V_o = \dfrac{h}{e}(\nu - \nu_{th})$$\implies \ V_o = \dfrac{6.6\times 10^{-34}}{1.6\times 10^{-19}}(8.2-3.3)\times 10^{14} = 2 \ V$

Question 10

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is :

Accepted Answer

Let $V_0$ be the stopping potentialIn a photoelectric effect, the maximum kinetic energy of electron , $K_{max}=eV_0$so, $V_0=\dfrac{K_{max}}{e}=\dfrac{0.5eV}{e}=0.5 V$    (as eV=1 electronic charge times 1 Volt)

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Photocell

Photo-Emissive Cell

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