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Questions:

Case Study Questions

4 x 4 = 16

When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic surface. This phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons depends on the wavelength of incident light and is independent of the intensity of light. Number of emitted photoelectrons depends on intensity. (hv - \(\phi\) is the maximum kinetic energy of emitted photoelectrons (where \(\phi\) is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference across the tube. Wavelength of characteristic X-ray depends on the atomic number.
(i) Einstein's photoelectric equation is

\(\text { (a) } E_{\max }=h v-\phi\)

\(\text { (b) } E=m c^{2}\)

\(\text { (c) } E^{2}=p^{2} c^{2}+m_{0}^{2} c^{4}\)

\(\text { (d) } E=\frac{1}{2} m v^{2}\)

(ii) Light of wavelength \(\lambda\) which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential will

(a) increase

(b) decrease

(c) be zero

(d) become exactly half

(iii) When ultraviolet rays incident on metal plate then photoelectric effect does not occur, it occur by incident of

(a) Infrared rays

(b) X-rays

(c) Radio wave

(d) Micro wave

(iv) If frequency (v > v₀) of incident light becomes n times the initial frequency (v), then K.E. of the emitted photoelectrons becomes (v₀ threshold frequency).

(a) n times of the initial kinetic energy

(b) More than n times of the initial kinetic energy

(c) Less than n times of the initial kinetic energy

(d) Kinetic energy of the emitted photoelectrons remains unchanged

(v) A monochromatic light is used in a photoelectric experiment. The stopping potential

(a) Is related to the mean wavelength	(b) Is related to the shortest wavelength
(c) Is not related to the minimum kinetic energy of emitted photoelectrons	(d) Intensity of incident light

According to Einstein, when a photon of light of frequency u or wavelength \(\lambda\) is incident on a photosensitive metal surface of work function \(\phi\)₀w' here \(\phi\)₀< hv (here, h is Planck's constant), then the emission of photoelectrons takes place. The maximum kinetic energy of the emitted photoelectrons is given by K_max = hv - \(\phi\)₀. If the frequency of the incident light is V₀called threshold frequency, the photoelectrons are emitted from metal without any kinetic energy. So hv₀ = \(\phi\)
(i) A metal of work function 3·3 eV is illuminated by light of wavelength 300 nm. The maximum kinetic energy of photoelectrons emitted is (taking h = 6·6 x 10^-34 Js)

(a) 0.413 eV

(b) 0.825 eV

(c) 1.65 eV

(d) 1.32 eV

(ii) The variation of maximum kinetic energy (K_max) of the emitted photoelectrons with frequency (v) of the incident radiations can be represented by

(iii) The variation of photoelectric current (i) with the intensity of the incident radiation (I) can be represented by

(iv) The graph between the stopping potential (V₀) and \(\left(\frac{1}{\lambda}\right)\) is shown in the figure \(\phi_{1}, \phi_{2}, \phi_{3}\) 3 are work function. Which of the following options is correct?

\(\text { (a) } \phi_{1}: \phi_{2}: \phi_{3}=1: 2: 3\)

\(\text { (b) } \phi_{1}: \phi_{2}: \phi_{3}=4: 2: 1\)

\(\text { (c) } \phi_{1}: \phi_{2}: \phi_{3}=1: 2: 4\)

(d) Ultraviolet light can be used to emit photoelectrons from metal 2 and metal 3 only

(v) Which of the following figures represent the variation of particle momentum and the associated de- Broglie wavelength?

A point source S of power 6.4 x 10^-3W emits mono energetic photons each of energy 6.0 eV. The source is located at a distance of 0·8 m from the centre of a stationary metallic sphere of work function 3·0 eV and of radius 1.6 x 10^-3 m as shown in figure. The sphere is isolated and initially neutral and photoelectrons are instantly taken away from sphere after emission. The efficiency of photoelectric emission is one for very 10⁵ incident photons.

(i) The'power received by the sphere through radiations is

\(\text { (a) } \frac{4 R^{2}}{P r}\)

\(\text { (b) } \frac{P R^{2}}{4 r^{2}}\)

\(\text { (c) } \frac{p^{2} R}{2 \pi r}\)

\(\text { (d) } \frac{P R}{4 r}\)

(ii) Number of photons striking the metal sphere per second is

(a) 6.7 x 10⁹

(b) 3.3 x 10⁹

(c) 6.7 x 10¹⁰

(b) 3.3 x 10¹⁰

(iii) The number of photoelectrons emitted per second is about

(a) 3.3 x 10⁴

(b) 6.7 x 10⁴

(c) 6.7 x 10¹⁵

(d) 3.3 x 10¹⁵

(iv) The photoelectric emission stops when the sphere acquires a potential about

(a) 2 V

(b) 3 V

(c) 4 V

(d) 6 V

(v) If the distance of source becomes double from the centre of the metal sphere then the power received by the sphere

\(\text { (a) } \frac{P R^{2}}{4 r^{2}}\)

\(\text { (b) } \frac{P R^{2}}{16 r^{2}}\)

\(\text { (c) } \frac{P R^{2}}{4 r}\)

\(\text { (d) } \frac{P^{2} R^{2}}{16 r^{2}}\)

When a monochromatic radiations of suitable frequency obtained from source S, after being filtered by a filter attached on the window W, fall on the photosensitive place C, the photo electrons are emitted from C, which get accelerated towards the plate A if it is kept at positive potential. These electrons flow in the outer circuit resulting in photoelectric current. Due to it, the micro ammeter shows a deflection. The reading of micrommeter measures the photoelectric current.

An experimental setup of verification of photoelectric effect is shown in figure. The voltage across the electrodes is measured with the help of an ideal voltmeter, and which can be varied by moving jockey J on the potentiometer wire. The battery used in potentiometer circuit is of 16 V and its internal resistance is 2 \(\Omega\) The resistance of 100 ern long po.t•e. ntiometer wire is 8 \(\Omega\).

The photocurrent is measured with the help of an ideal ammeter. Two plates of potassium oxide of area 50 cm² at separation 0.5 mm are used in the vacuum tube. Photocurrent in the circuit is very small, so we can treat the
potentiometer circuit as an independent circuit

Light	Violet	blue	Green	Yellow	Orange	Red
\(\lambda\) in \(\dot A\)	4000-5000	4500-5000	5000-5500	5500-6000	6000-6500	6500-7000

(i) When radiation falls on the cathode plate, a current of 2 \(\mu \)A is recorded in the ammeter. Assuming that the vacuum tube setup follows Ohm's law, the equivalent resistance of vacuum tube operating in the case when jockey is at end P is

(a) 8 x 10⁸\(\Omega\)

(b) 16 x 10⁶\(\Omega\)

(c) 8 x 10⁶\(\Omega\)

(d) 10 x 10⁶\(\Omega\)

(ii) It is found that ammeter current remains unchanged (2 \(\mu\)A) even when the jockey is moved from the end P to the middle point of the potentiometer wire. Assuming that all the incident photons eject electrons and the power of the light incident is 4 x 10^-6\(\Omega\)Then, the color of the incident light is

(a) Green

(b) Violet

(c) Red

(d) Orange

(iii) Which of the following colors may not give photoelectric effect for this cathode?

(a) Green

(b) Violet

(c) Red

(d) Orange

(iv) When other light falls on the anode plate, the ammeter reading zero till jockey is moved from the end P to the middle point of the wire PQ. Therefore, the deflection is recorded in the ammeter. The maximum kinetic energy of the emitted electron is

(a) 16 eV

(b) 8 eV

(c) 4 eV

(d) 10 eV

(v) If the intensity of incident radiation is increased twice, the number of photoelectrons emitted per second will be

(a) halves

(b) double

(c) remain same

(d) four times

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Answers Key:

Case Study Questions

4 x 4 = 16

(i) (a)
(ii) (a): According to Einstein's photoelectric equation,
\(e V_{0}=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}\)
As \(\lambda\)₀ is constant, so when \(\lambda\). is decreased, stopping potential (V₀) increases.
(iii) (b): It indicates that threshold frequency is greater than that of ultraviolet light. As X-rays have greater frequency than ultraviolet rays, so they can cause photoelectric effect.
(iv) (b): \(\mathrm{K.E}_{\cdot 1}=h \mathrm{v}-\phi\)
\({K.E}_{\cdot 2}=n h v-\phi=n(h v-\phi)+(n-1) \phi\)
\({K.E.} 2=n \mathrm{KE}_{1}+(n-1) \phi \)
\(\text { K.E. }_{2}>n \mathrm{KE}_{1}\)
(v) (b): Stopping potential is the measurement of maximum kinetic energy of emitted photoelectrons and kinetic energy of emitted photoelectrons is linearly related with the frequency of incident light corresponding (i.e., corresponding to shortest wavelength, KE. is maximum). Stopping potential is independent of intensity.

(i) (b): \(K_{\max }=h v-\phi_{0}=\frac{h c}{\lambda}-\phi_{0}\)
\(=\frac{\left(6 \cdot 6 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(300 \times 10^{-9}\right) \times\left(1 \cdot 6 \times 10^{-19}\right)}-3 \cdot 3\)
\(=4 \cdot 125 \times 3 \cdot 3=0 \cdot 825 \mathrm{eV}\)
(ii) (c) : \(K_{\max }=h v-\phi_{0}, \text { When } v=v_{0}, K_{\max }=0\)
\(\therefore \quad 0=h v_{0}-\phi_{0} \text { or } \phi_{0}=h v_{0}\)
If v < v₀ then K_max is negative, i.e., no photoelectric emission takes place. Thus, graph (c) is possible.
(iii) (a): Photoelectric current (i) is proportional to the intensity of the emission light. Thus, graph (a) is possible.
(iv) (c): From Einstein's photoelectric equation,
\(K_{\max }=e V_{0}=\frac{h c}{\lambda}-\phi\)
or \(V_{0}=\frac{h c}{e} \cdot \frac{1}{\lambda}-\frac{\phi}{e}\)
Graph of V₀ versus \(\frac{1}{\lambda}\) is a straight line Slope of straight line,\(\tan \theta=\frac{h c}{e}\)
\(\text { At } V_{0}=0, \text { we have }\)
\(\phi_{1}: \phi_{2}: \phi_{3}=\frac{h c}{\lambda_{01}}: \frac{h c}{\lambda_{02}}: \frac{h c}{\lambda_{03}}\)
\(0.001 h c: 0.002 h c: 0.004 h c\)
\(\therefore 1: 2: 4\)
(v) (d): de-Broglie wavelength
\(\lambda=\frac{h}{p} i . e ., \lambda \propto \frac{1}{p}\)
So the graph between \(\lambda\) and P is of the type shown is option (d).

(i) (b): Let R be the radius of the metallic sphere and r be its distance from the source S₀. The power received at the sphere is
\(P^{\prime}=\frac{P \times \pi R^{2}}{4 \pi r^{2}}=\frac{P R^{2}}{4 r^{2}}\)
(ii) (a): Number of photons striking the metal sphere per second is
\(n^{\prime}=\frac{P^{\prime}}{E}=\frac{6 \cdot 4 \times 10^{-9}}{6 \cdot 0 \times 1 \cdot 6 \times 10^{-19}}=6 \cdot 7 \times 10^{9} \mathrm{~s}^{-1}\)
(iii) (b): Number of photoelectrons emitted from metal sphere,\(\frac{n^{\prime}}{10^{5}}=\frac{6 \cdot 7 \times 10^{-9}}{10^{5}}=6 \cdot 7 \times 10^{4}\)
(iv) (b): Kinetic energy of the fastest photoelectrons is \(K_{\max }=6 \cdot 0-3 \cdot 0=3 \cdot 0 \mathrm{eV}\)
\(\therefore \text { Stopping potential, } V_{s}=\frac{K_{\max }}{e}=\frac{3 \cdot 0 \mathrm{eV}}{e}=3.0 \mathrm{~V}\)
(v) (b): When r = 2r, then power received by the sphere
\(P^{\prime \prime}=\frac{P \pi R^{2}}{4 \pi(2 r)^{2}}=\frac{1}{4}\left(\frac{P R^{2}}{4 r^{2}}\right)\)

(i) (c) :Equivalent resistance
\(R=\frac{V}{I}=\frac{16 \mathrm{~V}}{2 \times 10^{-6} \mathrm{~A}}=8 \times 10^{6} \Omega\)
(ii) (d): \(P=\frac{I h C}{e \lambda}\)
\(\lambda=\frac{\left(2 \times 10^{-6}\right)\left(6.6 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{\left(4 \times 10^{-6}\right)\left(1.6 \times 10^{-19}\right)}\)
\(=\frac{9.9}{1.6} \times 10^{-7} \mathrm{~m}=\frac{9900}{1.6} \dot A=6187 \dot A\)
Which is in the range of orange light
(iii) (c) : The range of wavelength for red light is beyond the wavelength of incident light.
(iv) (b): Stopping potential, \(V_{s}=8 \mathrm{~V}\)
and \(\mathrm{K.E.}=e V_{\mathrm{s}}\)
\(\therefore \quad \mathrm{K} . \mathrm{E} .=8 \mathrm{eV}\)
(v) (b): Photoelectric current varies linearly with the intensity of the incident radiation.