Q: The work function of caesium is 2.14 eV. Find a) the threshold frequency for caesium and b) the wavelength of the incident light if photocurrent is brought to zero by stopping potential of 0.60 V.

a) \$ W =hv\$Where \$W\$ = work function \$h\$ = planck constant \$v\$ = threshold frequency \$v= \dfrac{W}{h} =\dfrac{2.14×1.6×10^{-19}}{6.6×10^{-34}}= 5.2×10^{14}hertz\$b) Wavelength of incident photon =\$\dfrac{hc}{ev}+W\$By substituting the values we get Wavelength = \$4.5×10^{-7} m\$

Q: Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photo current becomes

When the frequency of incident light is halved of its original value i.e. \$1.5v_0\$ then it becomes less than the threshold value. In that case no photoelectric effect takes place. No photoelectrons would be emitted. The photoelectric current becomes zero.

Q: An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

For electronDe - Broglie wavelength \$\lambda_c = \dfrac{h}{p}\$where p is momentum \$p = mv\$also by energy we have \$E = \dfrac{1}{2} mv^2\$\$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}\$\$\Rightarrow p = \sqrt{2 mE}\$\$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}\$For photon energy \$\Rightarrow E = \dfrac{hc}{\lambda}\$\$\Rightarrow \lambda = \dfrac{h c}{E}\$\$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}\$\$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}\$option (A) is correct.

Q: The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is :

Given, the maximum kinetic energy: \$K_{max}=4 eV\$If \$V_0\$ be the stopping potential, then \$K_{max}=eV_0\$\$\Rightarrow eV_0=4 eV\$ \$\Rightarrow V_0=4V\$

Q: The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately:

If \$\lambda_0\$ be the threshold wavelength and \$\lambda\$ be the wavelength of incident light, the condition for photoelectron emission is \$\lambda<\lambda_0\$Thus, threshold wavelength is the longest wavelength for photoelectron emission. So, work function: \$W=\dfrac{hc}{\lambda_0}\$or, \$\lambda_0=\dfrac{hc}{W}=\dfrac{(6.62\times 10^{-34})(3\times 10^8)}{4\times 1.6\times 10^{-19}}=3.10\times 10^{-7} m=310\times 10^{-9} m=310 nm\$

Q: Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

The above graph is showing the variation of photoelectric current with the applied voltage. They have different intensities but same frequency.Here the upper one is for high intensity and lower one is for low intensity.

Q: The de Broglie wavelength of an electron with kinetic energy 120 eV is \$(Given \, h \, = \, 6.63 \, \times \, 10^{-34}J s, \, m_e \, = \, 9 \, \times \, 10^{-31} \, kg, \, 1 \, eV \, = \, 1.6 \, \times \, 10^{-19} J)\$:

As \$p =\sqrt{2m \ K.E.}\$ \$= \sqrt{2 \times 9 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}}\$ \$=5.88 \times 10^{-24}kg\, m \,s^{-1}\$de Broglie wavelength,\$\lambda = \dfrac{h}{p} = \dfrac{6.63\times 10^{-34}}{5.88 \times 10^{-24}} = 1.13 \times 10{-10}m = 1.13 \overset{0}{A}\$

Q: An electron is accelerated through a potential difference of 100 V. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

De Broglie wavelength, \$\lambda\left(=\dfrac{h}{p}\right)=\dfrac{h}{\sqrt{2meV}}\$\$=\dfrac{6.63\times10^{-24}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times100}}\$\$=1.227\times10^{-10}m=1.227A^0\$This wavelength belongs to X-ray spectrum.

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

(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 1

The work function of caesium is 2.14 eV. Find  a) the threshold frequency for caesium and b) the wavelength of the incident light if photocurrent is brought to zero by stopping potential of 0.60 V.

Accepted Answer

a) $ W =hv$Where $W$ = work function        $h$ = planck constant       $v$ = threshold frequency $v= \dfrac{W}{h} =\dfrac{2.14&#215;1.6&#215;10^{-19}}{6.6&#215;10^{-34}}= 5.2&#215;10^{14}hertz$b) Wavelength of incident photon =$\dfrac{hc}{ev}+W$By substituting the values we get Wavelength = $4.5&#215;10^{-7} m$

Question 2

A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

Accepted Answer

Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity $v$ is given by$\lambda = \dfrac {h}{mv} $       $ (\because p = mv)$de-Broglie wavelength of a proton of mass $m_{1}$ and kinetic energy $k$ is given by$\lambda_{1} = \dfrac {h}{\sqrt {2m_{1}k}}$           $(\because p = \sqrt {2mk})$$\lambda_1= \dfrac {h}{\sqrt {2m_{1}qV}} .... (i)$        $ [\because k = qV]$For an alpha particle mass $m_{2}$ carrying charge $q_{0}$ is accelerated through potential $V$, then$\lambda_{2} = \dfrac {h}{\sqrt {2m_{2}q_{0}V}}$$\because$ For $\alpha - particle\  \   (^{4}_{2}He)$ :  $ q_{0} = 2q$ and $m_{2} = 4m_{1}$$\therefore \lambda_{2} = \dfrac {h}{\sqrt {2\times 4m_{1}\times 2q\times V}} .... (ii)$The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get$\dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {h}{\sqrt {2m_{1}qV}} \times \dfrac {\sqrt {2\times m_{1} \times 4\times 2qV}}{h} = \dfrac {4}{\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}}$We get  $\dfrac{\lambda_1}{\lambda_2}= 2\sqrt {2}$

Question 3

Light of frequency 1.5 times the threshold frequency is incident on photosensitive material. If the frequency is halved and intensity is doubled, the photo current becomes

Accepted Answer

When the frequency of incident light is halved&#10;of its original value i.e. $1.5v_0$ then it becomes less than the threshold value. In&#10;that case no photoelectric effect takes place. No photoelectrons would be emitted.&#10;The photoelectric current becomes zero.

Question 4

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

Accepted Answer

For electronDe - Broglie wavelength $\lambda_c = \dfrac{h}{p}$where p is momentum $p = mv$also by energy we have $E = \dfrac{1}{2} mv^2$$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}$$\Rightarrow p = \sqrt{2 mE}$$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}$For photon energy $\Rightarrow E = \dfrac{hc}{\lambda}$$\Rightarrow \lambda = \dfrac{h c}{E}$$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}$$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}$option (A) is correct.

Question 5

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is :

Accepted Answer

Given, the maximum kinetic energy: $K_{max}=4 eV$If $V_0$ be the stopping potential, then $K_{max}=eV_0$$\Rightarrow eV_0=4 eV$ $\Rightarrow V_0=4V$

Question 6

The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately:

Accepted Answer

If $\lambda_0$ be the threshold wavelength and $\lambda$ be the wavelength of incident light, the condition for photoelectron emission is $\lambda<\lambda_0$Thus, threshold wavelength is the longest wavelength for photoelectron emission. So, work function: $W=\dfrac{hc}{\lambda_0}$or, $\lambda_0=\dfrac{hc}{W}=\dfrac{(6.62\times 10^{-34})(3\times 10^8)}{4\times 1.6\times 10^{-19}}=3.10\times 10^{-7} m=310\times 10^{-9} m=310 nm$

Question 7

Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

Accepted Answer

The above graph is showing the variation of photoelectric current with the applied voltage. They have different intensities but same frequency.Here the upper one is for high intensity and lower one is for low intensity.

Question 8

The de Broglie wavelength of an electron with kinetic energy 120 eV is $(Given \, h \, = \, 6.63 \, \times \, 10^{-34}J s, \, m_e \, = \, 9 \, \times \, 10^{-31} \, kg, \, 1 \, eV \, = \, 1.6 \, \times \, 10^{-19} J)$:

Accepted Answer

As $p =\sqrt{2m \ K.E.}$          $= \sqrt{2 \times 9 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}}$          $=5.88 \times 10^{-24}kg\, m \,s^{-1}$de Broglie wavelength,$\lambda = \dfrac{h}{p} = \dfrac{6.63\times 10^{-34}}{5.88 \times 10^{-24}} = 1.13 \times 10{-10}m = 1.13 \overset{0}{A}$

Question 9

An electron is accelerated through a potential difference of 100 V. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

Accepted Answer

De Broglie wavelength, $\lambda\left(=\dfrac{h}{p}\right)=\dfrac{h}{\sqrt{2meV}}$$=\dfrac{6.63\times10^{-24}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times100}}$$=1.227\times10^{-10}m=1.227A^0$This wavelength belongs to X-ray spectrum.

Question 10

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.