Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

In a Young double slit experiment, the distance between the slits is 50 $$\mu m$$ and the distance of the screen from the slits is 5 cm. An infrared monochromatic light is mood to produce an interference pattern with fringe width 1 mm. If the light source is replaced by an electron source to produce the same fringe width then the speed of the electrons is approximately

A
$$70 m/s$$

B
$$700 m/s$$

C
$$7000 m/s$$

D
$$70000 m/s$$

Solution

We know that :- $$w=\dfrac{\lambda D}{d}$$
Given :- $$d=50\times10^{-6}m, $$ $$w=5\times 10^{-2}m$$ and $$ w=10^{-3}m$$

$$\implies \lambda=\dfrac{wd}{D}=\dfrac{10^{-3}\times 50\times 10^{-6}}{5\times10^{-2}}m$$

$$\implies \lambda=10^{-6}m$$

For an electron to produce same fringe width, its de-Broglie wavelength should be same as $$\lambda$$.

Now,for electron:-
$$\lambda=\dfrac{h}{mv}$$

$$\implies v=\dfrac{h}{m\lambda}$$

$$\implies v=\dfrac{6.63\times 10^{-34}}{9.1\times10^{-31}\times 10^{-6}}m/s$$

$$\implies v\approx700m/s$$

Hence, answer is option-(B).
Question 2
1 / -0

A graph regarding photoelectric effect is shown between the maximum kinetic energy of electrons and the frequency of the incident light. On the basis of the data as shown in the graph, calculate the work function.

A
$$4\ eV$$

B
$$2\ eV$$

C
$$4.2\ eV$$

D
$$2.5\ eV$$

Solution

From the graph, threshold frequency
$$v_{0} = 10\times 10^{14}Hz$$
$$h = \dfrac {\triangle k_{max}}{\triangle v} = \dfrac {8\times 1.6\times 10^{-9}}{20\times 10^{14}} = 6.4\times 10^{-34}J$$
Thus, work function,
$$W = hv_{0} = 6.4\times 10^{-34} \times 10\times 10^{14}$$
$$= 4\times 1.6\times 10^{-19} = 4\ eV$$.
Question 3
1 / -0

A student performs an experiment on photoelectric effect using two materials A and B. A plot of stopping potential $$(V_0)$$ vs frequency $$(\upsilon)$$ is as shown in the figure.The value of h obtained from the experiment for both A and B respectively is (Given electric charge of an electron = 1.6 $$\times \, 10^{-19}$$ C)

A
$$3.2 \, \times \, 10^{-34}J s, \, 4 \, \times \, 10^{-34}J s$$

B
$$6.4 \, \times \, 10^{-34}J s, \, 8 \, \times \, 10^{-34}J s$$

C
$$1.2 \, \times \, 10^{-34}J s, \, 3.2 \, \times \, 10^{-34}J s$$

D
$$4.2 \, \times \, 10^{-34}J s, \, 5 \, \times \, 10^{-34}J s$$

Solution

$$K.E= hv-\phi$$
for stoping potential $$V_s=\dfrac{K.E}{e}$$
$$V_s=\dfrac{hv-\phi}{e}$$ ...........(1)

For metal A, Slope = $$\dfrac{h}{e} = \dfrac{2 - 0}{(10 - 5)} \times {10^{14}}$$

or h =$$ \dfrac{2 \times e}{5 \times 10^{14}} = \dfrac{2 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} = 6.4 \times 10^{-34} j s$$

For metal B , slope = $$\dfrac{h}{e} = \dfrac{2.5 - 0}{(15 - 10} \times 10^{14}$$

h =$$ \dfrac{2.5 \times e}{5 \times 10^{14}} = \dfrac{2.5 \times 1.6 \times 10^{-19}}{5 \times 10^{14}} = 8 \times 10^{-34} j s$$
Question 4
1 / -0

A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $$1.813 \, \times \, 10^{-4}$$. The mass of the particle is $$(m_e \, = \, 9.1 \, \times \, 10^{-31} \, kg)$$

A
$$1.67 \, \times \, 10^{-27} \, kg$$

B
$$1.67 \, \times \, 10^{-37} \, kg$$

C
$$1.67 \, \times \, 10^{-19} \, kg$$

D
$$1.67 \, \times \, 10^{-14} \, kg$$

Solution

Given : $$\dfrac{v}{v_e} = 3$$ and $$ \dfrac{\lambda}{\lambda_e} = 1.813 \times 10^{-4}$$

De Broglie wavelength of a moving particle having mass m and velocity v is given by
$$\because P=mv=\dfrac{h}{\lambda_D}$$

$$ h=mv \lambda _D=m_e v_e \lambda _{D_e}=mv \lambda _D $$

Mass of the particle, $$m = m_e \left[\dfrac{\lambda_{D_e}}{\lambda}\right]\left[\dfrac{v_e}{v}\right]$$

Substituting the value, we get
$$m = 9.1 \times 10^{-3} \times \dfrac{1}{1.813 \times 10^{-4}}\times \dfrac{1}{3}$$

or $$m = 1.67 \times 10^{-27}kg$$
Question 5
1 / -0

An electron (mass m) with an initial velocity $$\bar{v}\, = \,v_0\,\,\hat{i}(v_0 > 0)$$ is in an electric field $$ \bar{E} = -E_0\,\,\hat{i} $$ ($$E_0\, = \,constant \,>\, 0$$). Its de Broglie wavelength at time t is given by
Assume:$$\left ( \, \lambda_0 = \dfrac{h}{mv_0} \right ) $$

A
$$ \frac{\lambda_0}{\left ( 1\,+\,\dfrac{eE_0t}{mv_0} \right )} $$

B
$$ \lambda_0\left ( 1\,+\,\dfrac{eE_0t}{mv_0}\right ) $$

C
$$ \lambda_0 $$

D
$$ \lambda_0t $$

Solution

Here, $$ \bar{E} = -E_0\, \hat{i}$$ ; initial velocity $$\bar{v} = v_0 \,\, \hat{i} $$ Force action on electron due to electric field $$ \bar{F} = (-e)(-E_0 \, \hat{i}) = eE_0\, \hat{i} $$
Acceleration produced in the electron, $$ \bar{a} = \dfrac{\bar{F}}{m} = \dfrac{eE_0}{m} \, \hat{i}$$
Now, velocity of electron after time t, $$ \bar{v}_t = \bar{v} + \bar{a} \, t = \left ( v_0 + \dfrac{eE_0}{m} \right )\, \hat{i} \,\,\, or \,\,\, \left | \bar{v}_t \right |= v_0 + \dfrac{eE_0t}{m} $$

Now,
$$ \lambda_t = \dfrac{h}{mv_t}= \dfrac{h}{m\left ( v_0 + \dfrac{eE_0t}{m} \right )} = \dfrac{h}{mv_0\left ( 1 + \dfrac{eE_0t}{mv_0} \right )}$$

$$\lambda_t = \dfrac{\lambda_0}{\left ( 1 +\dfrac{eE_0t}{mv_0} \right )}$$ $$\left ( \because \, \lambda_0 = \dfrac{h}{mv_0} \right ) $$
Question 6
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A particle of mass $$M$$ at rest decays into two particle of masses $${m}_{1}$$ and $${m}_{2}$$ having non zero velocities. The ratio of the de Broglie wavelength of particles $${\lambda}_{1}/{\lambda}_{2}$$ :

A
$$\dfrac {{m}_{1}}{{m}_{2}}$$

B
$$\dfrac {{m}_{2}}{{m}_{1}}$$

C
$$1:1$$

D
$$\sqrt { \dfrac { { { m }_{ 2 } } }{ { { m }_{ 1 } } } } $$

Solution

Let velocity of particle of mass $$m_1$$ is $$v_1$$ and velocity of particle of mass $$m_2$$ is $$v_2$$.

There is no external force is acting so momentum will be conserved.

Before Decay, Linear momentum$$= 0$$,

After decay, Linear momentum$$= m_1v_1+m_2v_2$$

$$\Rightarrow m_1v_1+ m_2v_2=0$$ $$\Rightarrow m_1v_1=-m_2v_2$$

Magnitude $$m_1v_1=m_2v_2$$

de Broglie wavelength, $$\lambda=\dfrac{h}{mv}$$

$$\Rightarrow \dfrac{\lambda_1}{\lambda_2}=\dfrac{m_2v_2}{m_1v_1}=1$$

Hence, $$\lambda_1 : \lambda _2=1:1$$
Question 7
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An electromagnetic wave of wavelength $$\lambda$$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength $$\lambda'$$, then

A
$$\lambda \, = \, \dfrac{mc}{h} \, \lambda'^2$$

B
$$\lambda \, = \, \dfrac{3mc}{2h} \, \lambda'^2$$

C
$$\lambda \, = \, \dfrac{2mc}{h} \, \lambda'^2$$

D
$$\lambda \, = \, \dfrac{5mc}{h} \, \lambda'^2$$

Solution

Kinetic energy of emitted electron = Energy of incident photon
i.e. $$\dfrac{1}{2}mv^2 = h \nu$$

or $$\dfrac{p^2}{2m} = \dfrac{hc}{\lambda}$$

$$\left[ \because mv = p, \nu =\dfrac{c}{\lambda}\right]$$
or $$p = \sqrt{\dfrac{2mhc}{\lambda}}$$

de-Broglie wavelength of emitted electrons

$$\lambda' = \dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2mhc}{\lambda}}}$$ or

$$\lambda' = \sqrt{\dfrac{h\lambda}{2mc}}$$ $$\therefore \lambda = \dfrac{2mc}{h}\lambda'^2$$
Question 8
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An $$\alpha-$$particle and a proton have their masses in the ratio $$4\ :\ 1$$ and charges in the ratio $$2\ :\ 1$$. Find ratio of their de-Broglie wavelengths when both move with equal velocities.

A
$$1\ :\ 4$$

B
$$4\ :\ 1$$

C
$$1\ :\ 2$$

D
$$2\ :\ 1$$

Solution

We know,
De broglie wavelength= $$\lambda=\dfrac{h}{mv}$$
At constant velocity ($$v$$),
$$\lambda \propto \dfrac{1}{m}$$
$$\dfrac{\lambda_{\alpha}}{\lambda_{p}}=\dfrac{m_p}{m_{\alpha}}$$
Mass of $$\alpha$$ particle is 4 times mass of a proton.
$$\dfrac{m_p}{m_{\alpha}}=\dfrac{1}{4}$$
Hence,
$$\dfrac{\lambda_{\alpha}}{\lambda_{p}}=\dfrac{1}{4}$$
Hence option $$\textbf A$$ is correct.
Question 9
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The potential energy of particle of mass m varies as $$U (x) \, = \, \left\{\begin{matrix}E_0: \, \, \ 0 \, \leq \, x \, \leq \, 1 \\ 0: \, \ \, \, x \, > \, 1 \end{matrix}\right.$$ The de Broglie wavelength of the particle in the range $$0 \, \leq \, x \, \leq \, 1 \, is \, \lambda_1$$ and that in the range $$x \, > \, 1 \, is \, \lambda_2$$. If the total energy of the particle is $$2E_0$$, find $$\lambda_1,/\lambda_2.$$

A
$$\sqrt{2}$$

B
$$2$$

C
$$1$$

D
$$\dfrac{2}{\sqrt 2}$$

Solution

de Broglie wavelength of a particle of ///
$$\lambda = \dfrac{h}{\sqrt{2mK}}$$

where K is the kinetic energy of the particle.

For $$0 \leq x \leq 1, \, U(x) = E_0$$

As, total energy = Kinetic energy + potential energy
$$\therefore 2E_0 = K_1 + E_0$$
or $$K_1 = E_0$$
$$\therefore \lambda_1 = \dfrac{h}{\sqrt{2mK_1}} = \dfrac{h}{\sqrt{2mE_0}}$$

For $$ x > 1, U(x) = 0 \, \, \, \therefore K_2 = 2E_0$$
$$\therefore \lambda_2 = \dfrac{h}{\sqrt{2mK_2}} = \dfrac{h}{\sqrt{2m(2E_0)}}$$

Divide (i) by (ii), we get
$$\dfrac{\lambda_1}{\lambda_2} = \dfrac{h}{\sqrt{2mE_0}} \times \dfrac{\sqrt{2m(2E_0)}}{h} = \sqrt{2}$$
Question 10
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The collector of the photocell (in photoelectric experiment ) is made of tungsten while the emitter is of Platinum having work function of $$10 eV$$ Monochromatic radiation pf wavelength $$124 A $$ power $$100$$ watt is incident on emitter which enits photo electrons with a quantum efficiency of $$1 \% $$ . The accelerating voltage across the photocell is of $$ 10,000$$ volts (Use : $$hc = 12400 eV \mathring{A} $$ )

A
The power supplied by the accelerating voltage source is $$100 $$ watt.

B
The minimum wavelength of radiation coming from the tungsten target (collector) is $$ 1.23 \mathring{A} $$

C
The power supplied by the accelerating voltage source is $$10 $$ watt.

D
The minimum wavelength of radiation coming from the tungsten target (collector) is $$ 2.23 \mathring{A} $$

Solution

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Dual Nature of Radiation and Matter Test - 68

Result

Dual Nature of Radiation and Matter Test - 68

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Question 1

$$70 m/s$$

$$700 m/s$$

$$7000 m/s$$

$$70000 m/s$$

Solution

Question 2

$$4\ eV$$

$$2\ eV$$

$$4.2\ eV$$

$$2.5\ eV$$

Solution

Question 3

$$3.2 \, \times \, 10^{-34}J s, \, 4 \, \times \, 10^{-34}J s$$

$$6.4 \, \times \, 10^{-34}J s, \, 8 \, \times \, 10^{-34}J s$$

$$1.2 \, \times \, 10^{-34}J s, \, 3.2 \, \times \, 10^{-34}J s$$

$$4.2 \, \times \, 10^{-34}J s, \, 5 \, \times \, 10^{-34}J s$$

Solution

Question 4

$$1.67 \, \times \, 10^{-27} \, kg$$

$$1.67 \, \times \, 10^{-37} \, kg$$

$$1.67 \, \times \, 10^{-19} \, kg$$

$$1.67 \, \times \, 10^{-14} \, kg$$

Solution

Question 5

$$ \frac{\lambda_0}{\left ( 1\,+\,\dfrac{eE_0t}{mv_0} \right )} $$

$$ \lambda_0\left ( 1\,+\,\dfrac{eE_0t}{mv_0}\right ) $$

$$ \lambda_0 $$

$$ \lambda_0t $$

Solution

Question 6

$$\dfrac {{m}_{1}}{{m}_{2}}$$

$$\dfrac {{m}_{2}}{{m}_{1}}$$

$$1:1$$

$$\sqrt { \dfrac { { { m }_{ 2 } } }{ { { m }_{ 1 } } } } $$

Solution

Question 7

$$\lambda \, = \, \dfrac{mc}{h} \, \lambda'^2$$

$$\lambda \, = \, \dfrac{3mc}{2h} \, \lambda'^2$$

$$\lambda \, = \, \dfrac{2mc}{h} \, \lambda'^2$$

$$\lambda \, = \, \dfrac{5mc}{h} \, \lambda'^2$$

Solution

Question 8

$$1\ :\ 4$$

$$4\ :\ 1$$

$$1\ :\ 2$$

$$2\ :\ 1$$

Solution

Question 9

$$\sqrt{2}$$

$$2$$

$$1$$

$$\dfrac{2}{\sqrt 2}$$

Solution

Question 10

The power supplied by the accelerating voltage source is $$100 $$ watt.

The minimum wavelength of radiation coming from the tungsten target (collector) is $$ 1.23 \mathring{A} $$

The power supplied by the accelerating voltage source is $$10 $$ watt.

The minimum wavelength of radiation coming from the tungsten target (collector) is $$ 2.23 \mathring{A} $$

Solution

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