Dual Nature of Radiation and Matter Test

Result

Dual Nature of Radiation and Matter Test - 34

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The energy of a photon of light with wavelength $$5000 \overset {\circ}{A}$$ is approximately $$2.5\ eV$$. This way the energy of an X-ray photon with wavelength $$1\overset {\circ}{A}$$ would be:

A
$$\dfrac {2.5}{(5000)^{2}} eV$$

B
$$2.5\times 5000\ eV$$

C
$$\dfrac {25}{(5000)^{2}}\cdot eV$$

D
$$\dfrac {2.5}{5000} eV$$

Solution

Energy of photon, $$E = \dfrac {hc}{\lambda}$$
$$\Rightarrow 2.5\ eV = \dfrac {hc}{5000\times 10^{-10}}$$ (as $$\lambda = 5000 \overset {\circ}{A})$$

$$\Rightarrow hc = 2.5\times 5\times 10^{-7}\ eV$$

For X-ray photon, $$c = constant , h = constant$$

$$\Rightarrow E = \dfrac {hc}{\lambda_{x-ray}}$$

$$= \dfrac {2.5\times 5\times 10^{-7}}{1\times 10^{-10}} eV$$

$$= 2.5\times 5000\ eV$$

Hence, option $$B$$ is correct.
Question 2
1 / -0

When the kinetic energy of an electron is increased the wavelength of the associated wave will :

A
Increase

B
Decrease

C
Wavelength does not upon kinetic energy

D
None of the above

Solution

Since the kinetic energy of electron is given as $$E=\dfrac{p^2}{2m}$$
where $$p$$ is the momentum of the electron
$$\implies p=\sqrt{2mE}$$
Thus de-broglie wavelength=$$\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mE}}$$
Hence as energy of electron increases, the wavelength associated decreases.
Hence correct answer is option B.
Question 3
1 / -0

The energy of an electron of mass m moving with velocity V and de-Broglie wavelength $$\lambda$$ is __________. ('h' is Planck's constant)

A
$$\displaystyle\frac{h}{2m\lambda}$$

B
$$\displaystyle\frac{h^2}{2m\lambda^2}$$

C
$$\displaystyle\frac{h\lambda}{2m}$$

D
$$\displaystyle\frac{h}{m\lambda}$$

Solution

Since the electron is not moving in an electric field, its potential energy is zero and its total energy is equal to kinetic energy.
$$E=\dfrac { 1 }{ 2 } m{ V }^{ 2 }=\dfrac { 1 }{ 2 } m{ (\dfrac { h }{ m\lambda } ) }^{ 2 }=\dfrac { h^{ 2 } }{ 2m{ \lambda }^{ 2 } } $$
Question 4
1 / -0

If the uncertainty in the position of proton is $$\displaystyle 6\times { 10 }^{ -8 }m$$, then the minimum uncertainty in its speed will be:

A
$$\displaystyle 1\ { cms }^{ -1 }$$

B
$$\displaystyle 0.52\ { ms }^{ -1 }$$

C
$$\displaystyle 1\ { mms }^{ -1 }$$

D
$$\displaystyle 100\ { ms }^{ -1 }$$

Solution

According to heisenberg's uncertainty principle,
$$\Delta x \Delta p \geq \dfrac{h}{4 \pi}$$
$$\Delta x \times m\Delta v \geq \dfrac{h}{4 \pi}$$
$$\Delta v \geq \dfrac{h}{4 \pi m \Delta x} $$
Since, (Heigenberg's constant) $$ h = 6.626 \times 10^{-34} $$; (Mass of Proton) $$m = 1.67 \times 10^{-27} kg$$ and uncertainty in the position of proton is: $$ 6 \times 10^{-8} m$$
$$\Delta v \geq 0.52\ m/s $$
Question 5
1 / -0

Which of the following expression gives the de-Broglie relationship?

A
$$\displaystyle p=\frac { h }{ mv } $$

B
$$\displaystyle \lambda =\frac { h }{ mv } $$

C
$$\displaystyle \lambda =\frac { h }{ mp } $$

D
$$\displaystyle \lambda m=\frac { v }{ p } $$

Solution

The de-Broglie relation is,
$$\displaystyle \lambda =\frac { h }{ mv } $$
Where, $$\displaystyle \lambda $$ = de-Broglie wavelength
$$\displaystyle h$$=Planck's constant
$$\displaystyle m$$=mass of particle
and $$\displaystyle v$$= velocity of particle
Question 6
1 / -0

The de-Broglie wavelength of an electron moving with a velocity $$\dfrac{c}{2}$$ (where c is velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is:

A
$$1 : 4$$

B
$$1 : 2$$

C
$$1 : 1$$

D
$$2 : 1$$

Solution

De-broglie wavelength of the electron $$=\dfrac{h}{m_ev}=\dfrac{h}{m_e(c/2)}$$
Wavelength of photon $$=\lambda$$(say)
Thus, $$\lambda=\dfrac{h}{m_e(c/2)}$$
Ration of kinetic energies $$=\dfrac{\dfrac{1}{2}m_e(\dfrac{c}{2})^2}{h\dfrac{c}{\lambda}}=\dfrac{\dfrac{1}{2}m_e(\dfrac{c}{2})^2}{m_e\dfrac{c^2}{2}}=1:4$$
Question 7
1 / -0

The energy that should be added to an electron to reduce its de-Broglie wavelength from $$1$$nm to $$0.5$$nm is.

A
Four times the initial energy

B
Equal to the initial energy

C
Twice the initial energy

D
Thrice the initial energy

Solution

$$\lambda = \dfrac{h}{\sqrt {2mE}}$$
so $$\dfrac{\lambda'}{\lambda} = \dfrac{\sqrt E}{E'}$$
$$\therefore \dfrac{E}{E'} = (\dfrac{\lambda'}{\lambda})^2$$
$$ = (\dfrac{0.5}{1})^2 = \dfrac{1}{4}$$
so E' = 4E
so difference is 3E
Question 8
1 / -0

Because we cannot ever know the initial conditions of groups of objects with perfect precision, the later behaviour of complex systems of objects may be almost completely unpredictable, even the wonderfully successful equations derived from Newtonian mechanics.
The previous statement describes the essence of which theory ?

A
Newton's law of universal gravitation

B
Chaos theory

C
Relativity

D
Quantum Theory

E
Group Theory

Solution

The statement has the essence of quantum theory which postulates that both the position and momentum cannot be precisely measured at the same instant. This is called Uncertainty Principle. An associated error always exists in either or both. Hence behavior of complex systems becomes much more difficult to analyse and impossible to predict.
Question 9
1 / -0

From the pictured graph the value of Planck's constant can be determined to be about how much?

A
$$0.4\times 10^{-14}eV\cdot s$$

B
$$6.6\times 10^{-14}eV\cdot s$$

C
$$4.9 Hz$$

D
$$-2\ eV$$

E
$$6.6\times 10^{-34}eV\cdot s$$

Solution
Question 10
1 / -0

The energy of a photons is equal to the kinetic energy of a proton. If $$\lambda_1$$ is the de-Broglie wavelength of a proton, $$\lambda_2$$ the wavelength associated with the photon, and if the energy of the photon is E, then $$(\lambda_1/\lambda_2)$$ is proportional to:

A
$$E^4$$

B
$$E^{1/2}$$

C
$$E^2$$

D
$$E$$

Solution

Given : $$E_{proton} = E_{photon} =E$$
de-Broglie wavelength of the proton $$\lambda_1 = \dfrac{h}{p}$$ where $$p = \sqrt{2mE}$$
$$\implies$$ $$\lambda_1 = \dfrac{h}{\sqrt{2mE}}$$ .......(1)

Wavelength associated with photon is $$\lambda_2$$
$$\therefore$$ $$E = \dfrac{hc}{\lambda_2}$$ $$\implies \lambda_2 =\dfrac{hc}{E}$$ .........(2)

$$\therefore$$ $$\dfrac{\lambda_1}{\lambda_2} = \dfrac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}} = \dfrac{\sqrt{E}}{c\sqrt{2m}}$$
$$\implies$$ $$\dfrac{\lambda_1}{\lambda_2} \propto E^{1/2}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 34

Result

Dual Nature of Radiation and Matter Test - 34

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\dfrac {2.5}{(5000)^{2}} eV$$

$$2.5\times 5000\ eV$$

$$\dfrac {25}{(5000)^{2}}\cdot eV$$

$$\dfrac {2.5}{5000} eV$$

Solution

Question 2

Increase

Decrease

Wavelength does not upon kinetic energy

None of the above

Solution

Question 3

$$\displaystyle\frac{h}{2m\lambda}$$

$$\displaystyle\frac{h^2}{2m\lambda^2}$$

$$\displaystyle\frac{h\lambda}{2m}$$

$$\displaystyle\frac{h}{m\lambda}$$

Solution

Question 4

$$\displaystyle 1\ { cms }^{ -1 }$$

$$\displaystyle 0.52\ { ms }^{ -1 }$$

$$\displaystyle 1\ { mms }^{ -1 }$$

$$\displaystyle 100\ { ms }^{ -1 }$$

Solution

Question 5

$$\displaystyle p=\frac { h }{ mv } $$

$$\displaystyle \lambda =\frac { h }{ mv } $$

$$\displaystyle \lambda =\frac { h }{ mp } $$

$$\displaystyle \lambda m=\frac { v }{ p } $$

Solution

Question 6

$$1 : 4$$

$$1 : 2$$

$$1 : 1$$

$$2 : 1$$

Solution

Question 7

Four times the initial energy

Equal to the initial energy

Twice the initial energy

Thrice the initial energy

Solution

Question 8

Newton's law of universal gravitation

Chaos theory

Relativity

Quantum Theory

Group Theory

Solution

Question 9

$$0.4\times 10^{-14}eV\cdot s$$

$$6.6\times 10^{-14}eV\cdot s$$

$$4.9 Hz$$

$$-2\ eV$$

$$6.6\times 10^{-34}eV\cdot s$$

Solution

Question 10

$$E^4$$

$$E^{1/2}$$

$$E^2$$

$$E$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.