Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Electrons are emitted from a metal surface on incidence of green colour of light. Among the following group of colours, which group will emit electrons ?

A
Yellow , blue , red

B
Violet , red , yellow

C
Violet , blue , yellow

D
Violet , blue , sky blue

Solution

Work function $$ \phi = hv_0 = \dfrac{hc}{\lambda_0} $$

$$ \Rightarrow \, \phi \propto \dfrac{1}{\lambda_0} $$
They will be emitted at low wavelength
Hence, violet , blue , sky blue emit electron
Question 2
1 / -0

The wavelength of the de-Broglie wave associated with a thermal neutron of mass $$m$$ at absolute temperature $$T$$ is given by:
($$k$$ is the Boltzmann constant)

A
$$\displaystyle \frac{h}{\sqrt{mkT}}$$

B
$$\displaystyle \frac{h}{\sqrt{2mkT}}$$

C
$$\displaystyle \frac{h}{\sqrt{3mkT}}$$

D
$$\displaystyle \frac{h}{2\sqrt{mkT}}$$

Solution

The kinetic energy of the thermal neutron of mass m at absolute temperature T is given by

$$\dfrac {mv^2}{2}=kT$$

$$\Rightarrow p=mv=\sqrt {2mkT}$$

The wavelength of the deBroglie wave associated with the thermal neutron is

$$\lambda=h/p=\dfrac{h}{\sqrt {2mkT}}$$.
Question 3
1 / -0

The de-Broglie wavelength of a neutron at $$927^{0}$$C is$$\lambda$$. Its wavelength at $$27^{0}$$C is:

A
$$\dfrac{\lambda}{2}$$

B
$$\lambda$$

C
$$2\lambda$$

D
$$4\lambda$$

Solution

$$\lambda = \dfrac{h}{\sqrt{2mKT}}$$
$$T_1 = 927 + 273$$ , $$T_2 = 27 + 273$$
$$= 1200 \ K$$ , $$= 300\ K$$
$$\lambda \alpha \dfrac{1}{\sqrt{T}}$$
$$\therefore \dfrac{{\lambda}_1}{{\lambda}_2} = \dfrac{\sqrt{T_2}}{\sqrt{T_1}}$$
$$\dfrac{\lambda_1}{{\lambda}_2} = \dfrac{\sqrt{300}}{\sqrt{1200}}$$
$$\ {{\lambda}_2 = 2 \lambda}$$
Question 4
1 / -0

STATEMENT-1 : In photoelectric experiment the ejected electrons have same kinetic energy.
STATEMENT-2 : According to Einstein kinetic energy of every ejected electron is the difference in the energy of a photon and the work function of the metal.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True.

E
Statement-1 is False, Statement-2 is False.

Solution

When light of frequency f is incident on a metal surface that has a work function W, the maximum kinetic energy of the emitted electrons is given by:
$$KE_{max}=hf-\phi$$
Note that this is the maximum possible kinetic energy because $$\phi$$ is the minimum energy necessary to liberate an electron.Increasing the frequency of the incident beam, keeping the number of incident photons fixed increases the maximum kinetic energy of the photoelectrons emitted.So, Statement-1 is False, Statement-2 is False.
Question 5
1 / -0

Select correct alternative :
Statement -1 : An electric dipole can not produce zero electric field at a point which is situated at finite distance from the dipole.
Statement -2 : Mass of a moving photon is proportional to $$ \lambda^{-1} $$ .
Statement -3 : Equation of continuity for fluid is based on conservation of volume.

A
FFF

B
TTF

C
TFF

D
TTT

Solution

Statemen-1: The potential due to a electric dipole of moment (p) is:
$$\phi=\dfrac {1}{4\pi \epsilon} \dfrac{\vec{p}.\hat{r}}{r^2}$$
The electric field from a dipole can be found from the gradient of this potential which cannot be zero at a point which is situated at finite distance from the dipole.

Statemen-2: From de-Brogli formula: The mass of a moving electron is given by:
$$m=\dfrac{h}{\lambda v}$$. So, mass of a moving photon is proportional to $$\lambda^{-1}$$

Statemen-3: A continuity equation in physics is an equation that describes the transport of a conserved quantity. Since mass, energy, momentum, electric charge and other natural quantities are conserved under their respective appropriate conditions, a variety of physical phenomena may be described using continuity equations.

So the answer is TTF.
Question 6
1 / -0

A charge particle $$q_0$$ of mass $$m_0$$ is projected along the y-axis at $$t = 0$$ from origin with a velocity $$V_0.$$ If a uniform electric field $$E_0$$ also exists along the x-axis, then the time at which de Broglie wavelength of the particle becomes half of the initial value is :

A
$$\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

B
$$2\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

C
$$\sqrt3\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

D
$$3\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

Solution

Initial debrogle wavelength of the charge particle $$\lambda_0=\dfrac{h}{m_0v_0}$$
After any time t, $$\lambda'=\dfrac{h}{m_{0}\sqrt{V_{0}^{2}+\left ( \dfrac{qE_{0}t}{m_{0}} \right )^{2}}} $$
Suppose after time t, $$\lambda$$ becomes half of the initial value i.e. $$\lambda=\lambda_0 /2$$.
$$t=\sqrt3 \dfrac{m_0v_0}{q_0E_0}$$
Question 7
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If a strong diffraction peak is observed when electrons are incident at an angle `$$\mathrm{i}$$' from the normal to the crystal planes with distance `d' between them (see figure), de Broglie wavelength $$\lambda_{\mathrm{d}\mathrm{B}}$$ of electrons can be calculated by the relationship $$(\mathrm{n}$$ is an integer):

A
$$2\mathrm{d}$$ sin i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

B
$$\text{d}$$ cos i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

C
$$\mathrm{d}$$ sin i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

D
$$2\text{d}$$ cos i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

Solution
Question 8
1 / -0

A small 50 kg vehicle is designed to be moved in free space by a lamp which emits 100 watts of red light of $$\lambda = 6630 A^o$$. What is its acceleration?

A
$$6.66 \times 10^{-9}\ m/sec^{2}$$

B
$$3 \times 10^{-9}\ m/sec^{2}$$

C
$$2.22 \times 10^{-9}\ m/sec^{2}$$

D
$$4.44\ m/sec^{2}$$

Solution

In this problem it is the particle nature of light which determines the behavior of space vehicle. Since the emitted light carries away momentum and since the latter is carried, the rocket will be propelled to the left as shown in Figure.

Let N photons be emitted per sec.Each photon carries energy hv.
Then, power of lamp $$p=\dfrac{Energy\hspace{0.1in}of\hspace{0.1in}the\hspace{0.1in}photon \times number\hspace{0.1in}of\hspace{0.1in}photons}{time}=$$ $$Nhv$$
But $$v=\dfrac{c}{\lambda }$$
$$\therefore p=\dfrac{Nhc}{\lambda }$$ or
$$N=\dfrac{\rho \lambda }{hc}=\dfrac{100\times 6630\times 10^{-10}}{6.63\times 10^{-34}\times 3\times 10^{8}}$$
$$=\dfrac{663\times 10^{-7}}{663\times 10^{-28}\times 3}=\frac{1}{3}\times 10^{21}=3.33\times 10^{20}$$
Each particle will have momentum by de Broglie's hypothesis
$$\displaystyle \mathrm{p}=\frac{\mathrm{h}}{\lambda}$$
$$\dfrac{6.63\times 10^{-34}}{6630\times 10^{-10}}=10^{-27}$$ N-sec
Total force $$=\dfrac{d(n\phi )}{dt}$$ or $$\displaystyle \mathrm{F}=\mathrm{p}\frac{\mathrm{d}\mathrm{n}}{\mathrm{d}\mathrm{t}}=$$ p.N
$$\mathrm{F}=10^{-27}\times 3.33\times 10^{20}=3.33\times 10^{-7}$$ newton
Acceleration $$\dfrac{Force}{mass}=\dfrac{3.33\times 10^{-7}}{50}=6.66\times10^{-9}m/sec^{2}$$
Question 9
1 / -0

STATEMENT - 1 : In photoelectric experiment the ejected electrons have a wide range of kinetic energy even if the photons striking the surface have same energy.
STATEMENT - 2 : According to Einstein electrons can absorb any fraction of the energy of the striking photon.

A
Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.

B
Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True

Solution

According to Einstein, electron absorbs complete energy of photon but due
to its interaction with the atoms of material some energy is lost.
So, In photoelectric experiment the ejected electrons have a wide range of kinetic energy even if the photons striking the surface have same energy. This is true.
But the statement 'According to Einstein, electrons can absorb any fraction of the energy of the striking photon' is not correct. Because the electron can absorb the energy integral multiple of h$$\nu$$.
Question 10
1 / -0

If a hydrogen atom at rest, emits a photon of wavelength $$\lambda ,$$ the recoil speed of the atom of mass $$'m'$$ is given by

A
$$\dfrac{h}{m\lambda }$$

B
$$\dfrac{mh }{\lambda }$$

C
$$mh\lambda $$

D
none of these

Solution

From conservation of momentum,
$$P = const = \vec{p}_{photon}+\vec{p}_{atom}=0$$
Now, from de-broglie:
$$\lambda =\dfrac{h}{mv}\Rightarrow v=\dfrac{h}{m\lambda }$$
So, the answer is option (A).

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Dual Nature of Radiation and Matter Test - 60

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Dual Nature of Radiation and Matter Test - 60

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Question 1

Yellow , blue , red

Violet , red , yellow

Violet , blue , yellow

Violet , blue , sky blue

Solution

Question 2

$$\displaystyle \frac{h}{\sqrt{mkT}}$$

$$\displaystyle \frac{h}{\sqrt{2mkT}}$$

$$\displaystyle \frac{h}{\sqrt{3mkT}}$$

$$\displaystyle \frac{h}{2\sqrt{mkT}}$$

Solution

Question 3

$$\dfrac{\lambda}{2}$$

$$\lambda$$

$$2\lambda$$

$$4\lambda$$

Solution

Question 4

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True.

Statement-1 is False, Statement-2 is False.

Solution

Question 5

FFF

TTF

TFF

TTT

Solution

Question 6

$$\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

$$2\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

$$\sqrt3\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

$$3\dfrac{m_{0}v_{0}}{q_{0}E_{0}}$$

Solution

Question 7

$$2\mathrm{d}$$ sin i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

$$\text{d}$$ cos i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

$$\mathrm{d}$$ sin i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

$$2\text{d}$$ cos i $$=\mathrm{n}\lambda_{\mathrm{d}\mathrm{B}}$$

Solution

Question 8

$$6.66 \times 10^{-9}\ m/sec^{2}$$

$$3 \times 10^{-9}\ m/sec^{2}$$

$$2.22 \times 10^{-9}\ m/sec^{2}$$

$$4.44\ m/sec^{2}$$

Solution

Question 9

Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1.

Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

Solution

Question 10

$$\dfrac{h}{m\lambda }$$

$$\dfrac{mh }{\lambda }$$

$$mh\lambda $$

none of these

Solution

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