Physics Part Ii Chapter 11 Dual Nature Of Radiation And Matter

Infrared waves produced from hot bodies. Hence, it often called heat waves.

When electromagnetic waves hit the body the mass is lost by the momentum is conserved. i.e., transferred from Electromagnetic waves to the body.

A metal will show photoelectric emission in below situation:
E ≥ work function
Now according to the relation below:

$$\begin{gathered} \mathrm{E} = \mathrm{hv} = \frac{\mathrm{hc}}{\mathrm{\lambda }} \left(\mathrm{J}\right) \\ = \frac{\mathrm{hc}}{\mathrm{\lambda e}} \left(\mathrm{eV}\right) \\ = \frac{6.6 \mathrm{x} {10}^{-34} \mathrm{x} 3 \mathrm{x} {10}^8}{412.5 \mathrm{x} {10}^{-9} \mathrm{x} 1.6 \mathrm{x} {10}^{-19}} \\ \mathrm{E} = 3 \mathrm{ev} \end{gathered}$$

The work function of k and Na are less than the incident energy of the photon. Hence, only potassium (k) and Sodium (Na) will show photoelectric emission.

B.

1.41 eV

D.

d cos i = nλ_dB

D.

Diffraction pattern will be wider than the slit

D.

h v /c

C.

10−¹⁰ s

C.

1/m

C.

I₀

Maximum intensity is independent of slit width.

B.

five

For interference maxima, d sin θ = nλ Here d = 2λ
∴ sin θ = n/2 and is satisfied by 5 integral values of n (−2, −1, 0, 1, 2), as the maximum value of sin θ can only be 1.

C.

wavelength is halved and frequency remains unchanged

In vacuum, ε0 = 1
In medium, ε = 4
So, refractive index

Hence, it is clear that wavelength and velocity will become half but frequency remains unchanged when the wave is passing through any medium.

D.

is the same for all metals and independent of the intensity of the radiation.

C.

310 nm,

B.

25 μm

$$\begin{gathered} \sin {30}^o = \frac{\lambda }{b} \\ \Rightarrow \lambda = \frac{b}{2} = \frac{1 x {10}^{-6}}{2} = 5 x {10}^{-7} m \\ Fringe width, \omega = \frac{\lambda D}{d} \\ \Rightarrow d = \frac{\lambda D}{\omega } \\ = \frac{5 x {10}^{-7} x 0.5}{1 x {10}^{-2}} \\ d = 25 \mu m \end{gathered}$$

D.

1.78

A.

≈ 6 × 10⁵ ms^–1

B.

19 mm

$$\begin{gathered} Angular width = \frac{\lambda }{d} \\ So, 0.{20}^o = \frac{\lambda }{2 mm} \\ \Rightarrow \lambda = 0.{20}^o x 2 \\ Again, 0.{21}^o = \frac{\lambda }{d} \\ Now putting the value of \lambda \\ d = \frac{0.{20}^o x 2mm}{0.{21}^o} \\ d = 1.9 mm \end{gathered}$$

A.

1:2

Using Einstein's photoelectric equation,

E =W₀ + k_max

When light of frequency, 2v₀ is incident on a metal plate,

$$\begin{gathered} h\left(2v_o\right) = h{v}_o + \frac{1}{2}m{v}_1^2 \\ h{v}_0 = \frac{1}{2}m{v}_1^2 ..... \left(i\right) \end{gathered}$$

When light of frequency, 5v₀ is incident on a metal plate

h(5v₀) = hv₀  + $\frac{1}{2} m{v}_2^2$

$h{v}_0 = \frac{1}{2} m{v}_2^2 .... \left(ii\right)$

Dividing equ (i) and (ii)

$$\begin{gathered} \frac{1}{4} = \frac{v_1^2}{v_2^2} \\ \therefore \frac{v_1}{v_2} = \frac{1}{2} \end{gathered}$$

C.

$\frac{{\mathrm{b}}^2}{\mathrm{d}}, \frac{{\mathrm{b}}^2}{3\mathrm{d}}, \frac{{\mathrm{b}}^2}{5\mathrm{d}}$

The situation is shown in the figure,

The path difference between the waves reaching the point P is

$$\begin{gathered} ∆\mathrm{p} = \mathrm{BP} - \mathrm{AP} \\ = ({\mathrm{d}}^2 + {\mathrm{b}}^2{)}^{1/2} - \mathrm{d} = \mathrm{d}{\left(1 + \frac{{\mathrm{b}}^2}{{\mathrm{d}}^2}\right)}^{1/2} -\mathrm{d} \\ \approx \mathrm{d}\left[1 + \frac{1}{2}\frac{{\mathrm{b}}^2}{{\mathrm{d}}^2}\right] -\mathrm{d} = \frac{{\mathrm{b}}^2}{2\mathrm{d}} \\ \mathrm{For} \mathrm{a} \mathrm{dark} \mathrm{band} \\ ∆ \mathrm{p} = \frac{{\mathrm{b}}^2}{2\mathrm{d}} = (2\mathrm{n} -1)\frac{\mathrm{\lambda }}{2} \\ \mathrm{where}, \mathrm{n} = 1, 2, 3 \mathrm{or} \mathrm{\lambda } =\frac{1}{(2\mathrm{n}-1)}.\frac{{\mathrm{b}}^2}{\mathrm{d}} \\ \mathrm{Hence}, \mathrm{the} \mathrm{missing} \mathrm{wavelengths} \mathrm{are} \frac{{\mathrm{b}}^2}{\mathrm{d}}, \frac{{\mathrm{b}}^2}{3\mathrm{d}}, \frac{{\mathrm{b}}^2}{5\mathrm{d}} \end{gathered}$$

D.

$\frac{2\mathrm{\lambda mc}}{\mathrm{h}}$

The de-Broglie wavelength

$$\begin{gathered} \mathrm{h} = \frac{\mathrm{h}}{\mathrm{mv}} \\ \Rightarrow \mathrm{v} = \frac{\mathrm{h}}{\mathrm{m\lambda }} ... \left(\mathrm{i}\right) \end{gathered}$$

Energy of photon

${\mathrm{E}}_{\mathrm{p}} = \frac{\mathrm{hc}}{\mathrm{\lambda }} (\mathrm{since} \mathrm{\lambda } \mathrm{is} \mathrm{same})$ ..(ii)

Energy of photon

The kinetic energy of photon

$= \frac{{\mathrm{E}}_{\mathrm{p}}}{{\mathrm{E}}_{\mathrm{e}}} = \frac{\mathrm{hc}/\mathrm{\lambda }}{{\displaystyle \frac{1}{2}}{\mathrm{mu}}^2} = \frac{2\mathrm{hc}}{\mathrm{\lambda } {\mathrm{mv}}^2}$

Substituting value of c from Eq (i), we get

$\frac{{\mathrm{E}}_{\mathrm{p}}}{{\mathrm{E}}_{\mathrm{e}}} = \frac{2\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{m}} {\left({\displaystyle \frac{\mathrm{h}}{\mathrm{m\lambda }}}\right)}^2} = \frac{2\mathrm{\lambda mc}}{\mathrm{h}}$

B.

Two points propagating in two different parallel directions

We know that in any medium except vacuum or air, the velocities of different colours are different. Therefore, both red and green colours are refracted at different, angles of refraction. So, after emerging from glass slab through opposite parallel faces, they appear at two different points an move in two different parallel directions.

C.

${\left\{1 + \frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}\right\}}^{3/2}$

According to the question, the situation is shown in the figure below,

${\mathrm{E}}_1 = \frac{\mathrm{I}}{(\mathrm{LO}{)}^2} = \frac{\mathrm{I}}{{\mathrm{h}}^2} ... \left(\mathrm{i}\right)$

The illumination at the edge A is given by

$$\begin{gathered} {\mathrm{E}}_2 = \frac{\mathrm{I} \cos \mathrm{\theta }}{(\mathrm{LA}{)}^2} = \frac{\mathrm{I} \cos \mathrm{\theta }}{({\mathrm{h}}^2 + {\mathrm{r}}^2)} .... \left(\mathrm{ii}\right) \\ \mathrm{From} \mathrm{figure}, \cos \mathrm{\theta } = \frac{\mathrm{h}}{\sqrt{({\mathrm{h}}^2 + {\mathrm{r}}^2)}} \\ \therefore {\mathrm{E}}_2 = \frac{\mathrm{Ih}}{({\mathrm{h}}^2 + {\mathrm{r}}^2{)}^{3/2}} \\ \mathrm{Dividing} \mathrm{eq} \left(\mathrm{i}\right) \mathrm{by} \mathrm{eq} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2} = \frac{\mathrm{I}/{\mathrm{h}}^2}{\mathrm{Ih}/({\mathrm{h}}^2+{\mathrm{r}}^2{)}^{3/2}} = \frac{({\mathrm{h}}^2+ {\mathrm{r}}^2{)}^{3/2}}{{\mathrm{h}}^3} \\ = \frac{({\mathrm{h}}^2+ {\mathrm{r}}^2{)}^{3/2}}{{\mathrm{h}}^2} = {\left(1 + \frac{{\mathrm{r}}^2}{{\mathrm{h}}^2}\right)}^{3/2} \end{gathered}$$

D.

are one quarter as numerous

$\mathrm{I} \propto \left(\frac{1}{{\mathrm{d}}^2}\right)$

When the source is placed 2m away, then I' = I/4. The number of electrons emitted is directly proportional to intensity. Hence, the number of emitted electrons is reduced to one-fourth.

D.

2950 A^o

The minimum energy needed by an electron to come out from a metal surface is called the work function of the metal.

Suppose the cutoff wavelength is represented by λ_o.

$$\begin{gathered} \mathrm{So}, \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{o}}} =\mathrm{work} \mathrm{function}={\mathrm{W}}_{\mathrm{o}} \\ \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{o}}}=4.2 \times 1.6\times {10}^{-19} \\ {\mathrm{\lambda }}_{\mathrm{o}} =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{4.2\times 1.6\times {10}^{-19}} \\ =2946\times {10}^{-10} \mathrm{m} \\ {\mathrm{\lambda }}_{\mathrm{o}} =2950 {\mathrm{A}}^{\mathrm{o}} \end{gathered}$$

C.

5380 A^o

Here Work function 

W_o =23 eV =2.3 × 1.6 × 10^-19 J

Where λo is the wavelength

for sodium.

$$\begin{gathered} {\mathrm{W}}_{\mathrm{o}} =\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{o}}} \\ \Rightarrow {\mathrm{\lambda }}_{\mathrm{o} }=\frac{\mathrm{hc}}{{\mathrm{W}}_{\mathrm{o}}} \\ {\mathrm{\lambda }}_{\mathrm{o} }=\frac{6.6 \times {10}^{-34} \times 3 \times {10}^8}{2.3 \times 1.6 \times {10}^{-19}} \\ =5380 \times {10}^{-10} \mathrm{m} \\ =5380 {\mathrm{A}}^{\mathrm{o}} \end{gathered}$$

A.

30 kV

From the formula

$$\begin{gathered} \mathrm{e} \mathrm{V} =\mathrm{h}\frac{\mathrm{c}}{\mathrm{\lambda }} \\ \mathrm{V} =\frac{\mathrm{h} \mathrm{c}}{\mathrm{e} \mathrm{\lambda }} \\ =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 0.4125\times {10}^{-10}} \\ =30 \times {10}^3 \mathrm{volt} \\ =30 \mathrm{kvolt} \end{gathered}$$

C.

0.6 eV

The stopping potential depends on the kinetic energy of the electrons, which is affected only by the frequency of incoming light and not by its intensity.

The maximum kinetic energy of the photoelectron
(KE)_max = E - Φ  =1.8-1.2
            =  0.6  eV
i.e   eV_o = (KE)_max

or    eV_o = 0.6 eV

∴ Stopping potential V_o = 0.6 V

B.

4000 A^o

When the light ray passes through a glass slab, then new wavelength 

$$\begin{gathered} {\mathrm{\lambda }}^{\text{'}} = \frac{\mathrm{\lambda }}{\mathrm{\mu }} \\ =\frac{6000}{1.5} \\ = 4.5 {\mathrm{A}}^{\mathrm{o}} \end{gathered}$$

C.

A and B only

That metal will emit photoelectrons which has work function lower than that obtained with the radiation of 4100 A^o.

Work function for the wavelength of 4100 A^o is

$$\begin{gathered} \mathrm{W} = \frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }} \\ =\frac{6.62 \times {10}^{-14} \times 3 \times {10}^8}{4100 \times {10}^{-10}} \\ = 4.8 \times {10}^{-19} \mathrm{J} \\ = \frac{4.8 \times {10}^{-19}}{1.6 \times {10}^{-19}} \mathrm{eV} \\ \mathrm{W} = 3 \mathrm{eV} \end{gathered}$$

Now we have 

W_A = 1.92 eV

W_B = 2.0 eV

W_C = 5 eV

since W_A < W  and  W_B < W,  hence A and B will emit photoelectrons.

A.

${}_{34}\mathrm{Se}^{ 74}, {}_{31}\mathrm{Ga}^{ 71}$

The nuclei which have same number of neutrons but different atomic number and mass number are known as isotones. In choice (a) nuclei of ${}_{34}\mathrm{Se}^{74} \mathrm{and} {}_{31}\mathrm{Ga}^{71}$ are isotones as

A - z = 74 - 34 = 71 - 34 =71 - 31 = 40

A.

frequency

h - Planck's constant

E - Energy of photon

$$\begin{gathered} \mathrm{E} = \mathrm{h\nu } \\ \Rightarrow \left[\mathrm{h}\right] = \frac{\left[\mathrm{E}\right]}{\left[\mathrm{\nu }\right]} = \frac{\left[\mathrm{M} {\mathrm{L}}^2 \mathrm{T}{ }^{-2}\right]}{\left[{\mathrm{T}}^{ -1}\right]} \\ = \left[\mathrm{M} {\mathrm{L}}^2 {\mathrm{T}}^{ -1}\right] \end{gathered}$$

And I = moment of inertia = MR²

$$\begin{gathered} \Rightarrow \left[\mathrm{I}\right] = \left[\mathrm{M}\right] \left[{\mathrm{L}}^2\right] = \left[{\mathrm{ML}}^2\right] \\ \mathrm{Hence} , \\ \frac{\left[\mathrm{h}\right]}{\left[\mathrm{I}\right]} = \frac{\left[\mathrm{M} {\mathrm{L}}^2{\mathrm{T}}^{ -1}\right]}{\left[{\mathrm{T}}^{ -1}\right]} \\ = \left[{\mathrm{T}}^{ -1}\right] \\ = \frac{1}{\left[\mathrm{T}\right]}\Rightarrow \mathrm{dimension} \mathrm{of} \mathrm{frequency} \end{gathered}$$

Alternative method:-

$$\begin{gathered} \frac{\mathrm{h}}{\mathrm{I}} =\frac{{\displaystyle \raisebox{1ex}{$\mathrm{E}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\nu }$}\right.}}{\mathrm{I}} \\ = \frac{\mathrm{E} \times \mathrm{T}}{\mathrm{I}} =\frac{\left(\mathrm{kg} {\mathrm{m}}^2/{\mathrm{s}}^2\right){\displaystyle }{\displaystyle \times }{\displaystyle }{\displaystyle \mathrm{s}}}{\left(\mathrm{kg} {\mathrm{m}}^2\right)} \\ = \frac{1}{\mathrm{s}} = \frac{1}{\mathrm{time}} = \mathrm{freauency} \end{gathered}$$

Thus dimension of h/I is same as frequency.

D.

8 × 10⁶ m/s

The solution to our problem consists in Einstein's photoelectric equation. 

Einstein's photoelectric equation can be written as

$$\begin{gathered} \frac{1}{2} {\mathrm{mv}}^2 = \mathrm{h\nu } - \mathrm{\varphi } \\ \frac{1}{2}\mathrm{m}\times {\left(4 \times {10}^6\right)}^2 = 2{\mathrm{h\nu }}_{\mathrm{o}} - {\mathrm{h\nu }}_{\mathrm{o}} ...\left(\mathrm{i}\right) \\ \mathrm{and} \\ \frac{1}{2}\times \mathrm{m} \times {\mathrm{v}}^2 = 5 {\mathrm{h\nu }}_{\mathrm{o}}- {\mathrm{h\nu }}_{\mathrm{o}} ......\left(\mathrm{ii}\right) \end{gathered}$$

Dividing equation (ii) by eq (i)

$$\begin{gathered} \Rightarrow \frac{{\mathrm{\nu }}^2}{{\left(4 \times {10}^6\right)}^2} = \frac{4{\mathrm{h\nu }}_{\mathrm{o}}}{{\mathrm{h\nu }}_{\mathrm{o}}} \\ \Rightarrow {\mathrm{\nu }}^2 = 4 \times 16 \times {10}^{12} \\ \therefore {\mathrm{v}}^2 = 64 \times {10}^{12} \\ \mathrm{\nu } = 8 \times {10}^6 \mathrm{m}/\mathrm{s} \end{gathered}$$

The efficiency of photoelectric effect is less than 1 o .ie, number of photons less than 1 % are capable of ejecting photoelectrons.

A.

< 124 kV

From conservation of energy the kinetic energy of electron equals the maximum photon energy (we neglect the he work function $\mathrm{\varphi }$ because it is normally so small compared to eV_o ). 

       eV_o = h v_max

         $$\begin{gathered} {\mathrm{eV}}_{\mathrm{o}} = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\min }} \\ \therefore {\mathrm{V}}_{\mathrm{o}} = \frac{\mathrm{hc}}{{\mathrm{e\lambda }}_{\min }} \\ \Rightarrow {\mathrm{V}}_{\mathrm{o}} = \frac{12400 \times {10}^{-10}}{{10}^{-11}} \end{gathered}$$

                 = 124 kV

Hence, accelerating voltage for electrons in X-ray machine should be less than 124 kV.

B.

frequency

In photoelectric effect for a given photosensitive material, there exists a certain minimum cut-off frequency, called the threshold frequency, below which no emission of photoelectrons takes place no matter how intense the light is.

B.

black

If an object reflects the colour of light incident on it, it will appear with that colour but if object absorbs the colour of light, it will appear to be black. since the given wavelength does not belong to green, so it will be absorbed by the leaf and hence, it would appear to be black.

D.

colour changes to red and also intensity gets reduced

As batteries wear out, temperature of filament of flash light attains lesser value, therefore intensity of radiation reduces. Also dominating wavelength (λ_m ) in spectrum, which is the red colour, increases.

D.

None all have same wavelength

On the basis of dual nature of light, Lious de-Broglie suggested that the dual nature is not only of light, but each moving material particle has the dual nature.

He assumed a wave to be associated with each moving material particle which is called the matter wave. The wavelength of this wave is determined by the momentum of the particle. If p is the momentum of the particle, the wavelength of the wave associated with it is

                                     $\mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}}$ 

Where h is Planck's constant.

Since it is given that alpha, beta and gamma rays carry same momentum, so they will have same wavelength.

A.

1.6 × 10^-17 J

The energy transferred to electrons is e V joules. So each electron gains kinetic energy equal to the amount of energy transferred from the electrical supply. The electron starts from rest ( near enough ) so the Kinetic energy gained is given by 1 / 2 mv² where m is the mass of electron and v is speed.

KE = e V

    = 1.6 × 10^-19 × 100

KE = 1.6 × 10^-17 J

A.

ν - ν_o

When light falls on a metallic surface, ejection of photoelectron results. In this process, conservation
of energy holds.

Thus, from the  law of conservation of energy,

Energy imparted by the photon 

             = Maximum kinetic energy of the emitted electron + work function of the metal

⇒   hv = (KE)_max  + $\mathrm{\varphi }$

but $\mathrm{\varphi } = {\mathrm{h\nu }}_{\mathrm{o}}$ being threshold frequency.

∴          ${\left(\mathrm{KE}\right)}_{\max } = \mathrm{hv} - {\mathrm{hv}}_{\mathrm{o}}$

⇒          (KE)_max ∝ v - v_o

D.

According to photoelectric equation

           E_k = hv - hv_o               ....... (i)

If the energy of photon ( hv ) is less than the work function ( hv_o ) of metallic surface, then
electrons will never be ejected from surface regardless of intensity of incident light.

Also from equation (i) when v = v_o , E_k = 0

∴ Graph (d) represents variation of E_k with v.

B.

$\frac{\mathrm{d} \left(\mathrm{n} + \sqrt{2}\right)}{\mathrm{n} \sqrt{2}}$

The depth of the image of an object submerged in a transperent medium; it is reduced from the real depth of the object by a factor equal to the relative index of refraction of the medium with respect to air.

Refractive index μ = $\frac{\mathrm{Real} \mathrm{depth} \left(\mathrm{d}\right)}{\mathrm{Apparent} \mathrm{depth} \left(\mathrm{x}\right)}$

For 1st liquid 

             $\sqrt{2} = \frac{\mathrm{d}}{{\mathrm{x}}_1}$

⇒            x₁ = $\frac{\mathrm{d}}{\sqrt{2}}$

Similarly, for 2nd liquid,

               n = $\frac{\mathrm{d}}{{\mathrm{x}}_2}$

⇒             x₂ = $\frac{\mathrm{d}}{\mathrm{n}}$

Total apparent depth = x₁  + x₂

                               = $\frac{\mathrm{d}}{\sqrt{2}} + \frac{\mathrm{d}}{\mathrm{n}}$

                                = $\frac{\mathrm{d} \left(\mathrm{n} + \sqrt{2}\right)}{\mathrm{n} \sqrt{2}}$

The figure shows the apparent depth for (a) normal and (b) oblique viewing

A.

wave theory of light

Light's speed in vaccum is the fastest it can go. However in air and glass photons occasionally bounce off other particles so the photons do not travel in a straight line; they have travel further than they would in vacuum so they appear to travel slower.

 Wave theory of light predicted that light could reflect off shiny surfaces, react or bend when moving from the material into another, and diffract (or spread ) around objects or moving through slits.

Light travels faster in air than that in glass. This is accordance with wave theory of light.

C.

A and B only

Work function for wavelength of 4100 A^o is

          W = $\frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }}$ 

          W = $\frac{6.63 \times {10}^{-34} \times 3 \times {10}^8}{4100 \times {10}^{-10}}$

          W = 4.8 × 10^-19 J

          W = $\frac{4.8 \times {10}^{-19} }{1.6 \times {10}^{-19}}$ eV

          W = 3 eV

Now we have,

        W_A = 1.22 eV

        W_B = 2.0 eV

        W_C = 5 eV

Since W_A < W and W_B < W hence A and B will emit photoelectrons. 

So only metals having a work function less than 3 eV can emit photoelectrons for the incident of wavelength 4100 A^o.

B.

$\sqrt{3} \mathrm{l}$

Ratio of intensities  $\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} = \frac{\cos {\displaystyle }{\displaystyle {\mathrm{\theta }}_1}}{\cos {\displaystyle }{\displaystyle {\mathrm{\theta }}_2}}$

                                 = $\frac{\cos {60}^{\mathrm{o}}}{\cos {30}^{\mathrm{o}}}$

                           ${\mathrm{I}}_2 = \sqrt{3} {\mathrm{I}}_1$

A.

λ_ph > λ_e

According to DeBroglie Wavelength

Wavelength λ =  $\frac{\mathrm{hc}}{\mathrm{p}}$

                      = $\frac{\mathrm{hc}}{\sqrt{2 \mathrm{m} \mathrm{E}}}$

∴                λ ∝ $\frac{1}{\sqrt{\mathrm{m}}}$

Mass of photon is less than that of electron.

 Wavelength of photon is greater than that of electron λ_ph > λ_e

D.

344.4 A^o

Work function W = $\frac{\mathrm{hc}}{\mathrm{\lambda }}$

 h is  Planck's constant

 λ  is wavelength 

 c is the speed of light in vacuum

⇒              $\mathrm{\lambda } = \frac{\mathrm{hc}}{\mathrm{W}}$

⇒                 = $\frac{6.6 \times {10}^{-34} \times 3 \times {10}^8}{5.26 \times {10}^{-18}}$

 ⇒            λ  = 344.4 A^o

C.

intensity of incident beam

By increasing the intensity of incident beam, the number of emitted photo-electrons can be increased. Hence, number of photo-electrons emitted is proportional to the intensity of the incident beam.

A.

(i), (iii)

According to Einstein photoelectric equation,

       E = K_max + $\mathrm{\varphi }$

        $\frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }}$ = $\mathrm{\varphi }$+ eV_s

∴     V_s = $\left[ \frac{\mathrm{hc}}{\mathrm{\lambda }} - \mathrm{\varphi } \right]$ $\frac{1}{\mathrm{e}}$

   V_s versus $\frac{1}{\mathrm{\lambda }}$ graph is a straight line.

    Slope 

           $\mathrm{tan\theta } = \frac{\mathrm{h} \mathrm{c}}{\mathrm{e}}$

  The curves intersect the $\frac{1}{\mathrm{\lambda }}$ axis, where V_s is zero

∴   $\frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }} - \mathrm{\varphi }$ = 0

⇒   $\frac{1}{\mathrm{\lambda }}$ = $\frac{\mathrm{\varphi }}{\mathrm{h} \mathrm{c}}$

For metal 1

      $\frac{{\mathrm{\varphi }}_1}{\mathrm{h} \mathrm{c}} = 0.001$

For metal 2

       $\frac{{\mathrm{\varphi }}_2}{\mathrm{hc}}$ = 0.002

For metal 3

         $\frac{{\mathrm{\varphi }}_3}{\mathrm{hc}}$ = 0.004

∴   ${\mathrm{\varphi }}_1 : {\mathrm{\varphi }}_2 : {\mathrm{\varphi }}_3$ = 1 : 2 : 4

For metal 1

    ${\mathrm{\lambda }}_{{\mathrm{O}}_1} = \frac{\mathrm{hc}}{\mathrm{\varphi }} = \frac{1}{0.001}$ = 1000 nm

For metal 2, threshold wavelength,

     ${\mathrm{\lambda }}_{{\mathrm{O}}_2} = \frac{\mathrm{hc}}{\mathrm{\varphi }}$ = $\frac{1}{0.002}$ = 500 nm

Simillarly, 

      ${\mathrm{\lambda }}_{{\mathrm{O}}_3} = \frac{\mathrm{hc}}{\mathrm{\varphi }} = \frac{1}{0.004}$ = 250 nm

          λ_voilet = 400 nm

 For metal 1     

    λ_voilet < λ_threshold

  For metal 3 

           λ_voilet >  λ_threshold

B.

2.23 x 10¹⁷ / m² s

Given:-

    λ = 500 nm = 500 x 10^-9 m

Radiation power P = 10 W

As     P = n_o $\frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }}$

         n_o = $\frac{\mathrm{P} \mathrm{\lambda }}{\mathrm{hc}}$                            ......(i)

where, n_o is number of photons per second.

At distance r from point source, number of photons/area/time

       n' = $\frac{{\mathrm{n}}_{\mathrm{o}}}{4\mathrm{\pi } {\mathrm{r}}^2}$

       n' = $\frac{\mathrm{P} \mathrm{\lambda }}{\mathrm{hc}. 4{\mathrm{\pi r}}^2}$                         [ from eq. (i) ]

           = $\frac{10 \times 500 \times {10}^{-9}}{6.6 \times {10}^{-34} \times 3 \times {10}^8 \times 4\mathrm{\pi } {\left(3\right)}^2}$

           = 2.23 × 10¹⁷ m/s

D.

1 : 2

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is

         K_max = h ν  $- {\mathrm{\varphi }}_0$

where hu is the energy of incident photon and ${\mathrm{\varphi }}_0$ is the work function.

    K_max = hv $- {\mathrm{\varphi }}_0$

∴   $\frac{1}{2} {\mathrm{mv}}_{\max }^2$ = hv $- {\mathrm{\varphi }}_0$

As per question

     $\frac{1}{2} {\mathrm{mv}}_{\max }^2$= 1eV $-$ 0.5 eV

                  = 0.5 eV               ....(i)

and $\frac{1}{2} {\mathrm{mv}}_{\max }^2 = 2.5 \mathrm{eV} - 0.5 \mathrm{eV}$

                  = 2 eV                ......(ii)

Dividing eqn. (i) by eqn. (ii), we get

     $\frac{{\mathrm{v}}_{{\max }_1}^2}{{\mathrm{v}}_{{\max }_2}^2} = \frac{0.5 \mathrm{eV}}{2 \mathrm{eV}}$

              = $\frac{1}{4}$

   $\frac{{\mathrm{v}}_{{\max }_1}}{{\mathrm{v}}_{{\max }_2}} = \sqrt{\frac{1}{4}}$

   $\frac{{\mathrm{v}}_{{\max }_1}}{{\mathrm{v}}_{{\max }_2}} = \frac{1}{2}$

C.

If assertion is true but reason is false.

According to photoelectric equation,

   eV_s = hv $-$ W

        V_s = $\left(\frac{\mathrm{h}}{\mathrm{e}}\right) \mathrm{\nu } - \left(\frac{\mathrm{W}}{\mathrm{e}}\right)$

h = Planck's constant

For V_s versus v graph

     slope = $\frac{\mathrm{h}}{\mathrm{e}}$ = constant

D.

β-particle

de Broglie wavelength

             λ = $\frac{\mathrm{h}}{\mathrm{mv}}$

⇒           λ  ∝ $\frac{1}{\mathrm{m}}$                 (where v is constant)

⇒          m_{$\mathrm{\alpha }$} > m_n  > m_p  > m_β

∴            λ_{$\mathrm{\alpha }$} < λ_n < λ_p < λ_β 

Therefore, β-particle has the longest de Broglie wavelength.

B.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Photoelectric emission have been adequately explained by Einstein on the basis of photon theory of light.

B.

6.63 × 10^-34 m

The de-Broglie equation is an equation used to describe the wave properties matter, specifically the wave nature of the electron as given below           

de-Broglie wavelength is given by

          $\mathrm{\lambda } =\frac{\mathrm{h}}{\mathrm{mv}} = \frac{\mathrm{h}}{\mathrm{p}}$  

Where  λ  is wavelength

           h - Plankck's constant

          λ  = $\frac{\mathrm{h}}{\sqrt{2 \mathrm{m} \mathrm{K}}}$

Here m = 1 kg, K = 0.5 J

         h = 6.63 × 10^-34 Js

                Find λ = ?

               λ = $\frac{6.63 \times {10}^{-34}}{\sqrt{2 \times 1 \times 0.5}}$

               λ = 6.63 × 10^-34 m

D.

all the above choices are correct

The three experimental facts listed under (a), (b) and (c) cannot be explained on the basis of wave theory of light. All these facts support the quantum (photon) nature of light.