Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

In case of electrons and photons having the same wavelength. What is same for them ?

A
Energy

B
Velocity

C
Momentum

D
Angular momentum

Solution

For electron, Debroglie wavelength
$$ \lambda _e = \cfrac{h}{p}$$
For Photon,
$$ \lambda _p = \cfrac{h}{p}$$
Hence equating, momentum has to be same for both of them to have same wavelength
Question 2
1 / -0

Directions For Questions

A physicist wishes to eject electrons by shining light on a metal surface. The light source emits light of wavelength of 450 nm. The tables lists the only available metals and their work functions.
Metal $$W_0(eV)$$
Barium 2.5
Lithium 2.3
Tantalum 4.2
Tungsten 4.5

...view full instructions

Which metal(s) can be used to produce electrons by the photoelectric effect from given source of light ?

A
Barium only

B
Barium or lithium

C
Lithium, tantalum or tungsten

D
Tungsten or tantalum

Solution

Photoelectric effect
$$KE_{max} = hv - W_o$$
The minimum energy to knock out electrons
$$ KE \ge 0$$
$$ hv > W_o$$
$$\cfrac{hc}{\lambda} >W_o$$ ------------------(1)
Here, available energy = $$E = \cfrac{hc}{\lambda} = 2.75 eV$$
For Photoelectric effect
$$ W_o < 2.75 eV$$ by (1)
From the table, only true for Ba and Li
Question 3
1 / -0

The work function of aluminium is 4.2 eV. If two photons, each of energy 3.5 eV strike an electrons of aluminium, then emission of electrons

A
Will be possible

B
Will not be possible

C
Data is incomplete

D
Depends upon the density of the surface

Solution

$$ Photo electric \ \ effect$$
$$ KE = hv- W_o$$
for electron emission
$$KE>0$$
$$hv > W_o$$
given
$$hv =3.5 eV$$
$$ hv< W_o$$
hence no electron is emitted.
Note: photons are packets of energy and the energy required to knock out electrons is to be provided all at once. It cannot absorb energy from more photons in steps and come out.
Question 4
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Each question contains statement 1 and statement 2. Choose the correct answer (only one option is correct) from the following options.

Statement-1 : The de-Broglie wavelength of a molecule (in a sample of ideal gas) varies inversely as the square root of absolute temperature.
Statement-2 : The rms velocity of a molecule (in a sample of ideal gas) depends on temperature.

A
Statement-1 is false, Statement -2 is true

B
Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for statement-1

C
Statement -1 is true, Statement-2 is true; Statement -2 is not correct explanation for statement-1

D
Statement-1 is true, Statement -2 is false

Solution

$$v_{rms} = \sqrt{\dfrac{3RT}{M}}$$

$$v_{rms} \propto \sqrt{T}$$

De broglie wavelength
$$ \lambda = \cfrac{h}{mv}$$
$$ \lambda \propto \cfrac{1}{v_{rms}}$$
$$ \lambda \propto \cfrac{1}{\sqrt{T}}$$
Question 5
1 / -0

If 5% of the energy supplied to a bulb is radiated as visible light, the number of visible quanta emitted per second by a 100W bullb, assuming the wavelength of visible light to be $$5.6 \times 10^{-5}cm$$, is

A
$$1.4 \times 10^{19}$$

B
$$1.4 \times 10^{20}$$

C
$$2 \times 10^{19}$$

D
$$2 \times 10^{20}$$

Solution

Energy of single photon = $$\cfrac{hc}{\lambda} $$

If there are $$n$$ photons, total energy = $$ \cfrac{nhc}{\lambda}$$

Total energy emitted =$$ 0.05 \times 100 W = 5 W/s$$
Equating

$$ \cfrac{nhc}{\lambda} = 5$$

$$ n = \cfrac{5 \lambda}{hc}$$

Substitute values

$$n = \cfrac{5 \times 5.6 \times 10^{-7}}{6.26 \times 10^{-34} \times (3 \times 10^8)}$$

$$n= 1.4 \times 10^{19}$$
Question 6
1 / -0

A material particle with a rest mass $$'m_0\ '$$ is moving with speed of light 'c'. The de-Broglie wavelength associated is given by

A
$$\dfrac{h}{m_0c}$$

B
$$\dfrac{m_0c}{h}$$

C
Zero

D
$$\infty $$

Solution

Theory of special relativity

$$ m' = \gamma m = \cfrac{m}{\sqrt{1 -\cfrac{V^2}{c^2}}}$$

Debroglie wavelength

$$ \lambda = \cfrac{h}{m'v} = \cfrac{h \sqrt{1 -\cfrac{V^2}{c^2}}}{mv}$$

If $$v=c$$
then $$ \lambda =0$$
Question 7
1 / -0

A monochromatic source of light operating at 200 W emits $$4 \times 10^{20}$$ photons per second. Find the wavelength of light.

A
$$400 nm$$

B
$$200 nm$$

C
$$4 \times 10 ^{-10} \mathring { A } $$

D
None of these

Solution

Energy of single photon = $$\cfrac{hc}{\lambda} $$
If there are $$n$$ photons, total energy = $$ \cfrac{nhc}{\lambda}$$
Total energy emitted =$$ 200 W$$
Equating

$$ \cfrac{nhc}{\lambda} = 200$$

$$ n = \cfrac{200 \lambda}{hc}$$

Substitute values: $$4 \times 10^{20} = \cfrac{200 \times \lambda}{6.26 \times 10^{-34} \times (3 \times 10^8)}$$

$$\lambda = 400 nm$$
Question 8
1 / -0

Calculate the de Broglie wavelength for a beam of electron whose energy is 100 eV:

A
1 A

B
1.23 A

C
2.46 A

D
None of these

Solution

As we know,
E = $$\displaystyle\frac{1}{2}$$mv$$^2$$ = 100 eV = 100 $$\times$$ 1.6 $$\times$$ 10$$^{-19}$$ J

v$$^2$$ = $$\displaystyle\frac{2 E}{m}$$

v = ($$\displaystyle\frac{2 E}{m})^{1/2)}$$

$$\lambda$$ = $$\displaystyle\frac{h}{mv}$$ = $$\displaystyle\frac{h}{\sqrt{2mE}}$$ metre

$$\lambda$$ = $$\displaystyle\frac{6.6 \times 10^{-34}}{(2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}^{1/2}$$

=1.23 $$\times$$ 10$$^{-10}$$ m =$$ 1.23 A^0$$
Question 9
1 / -0

If velocity of a particle A is 50% of velocity of particle B and mass of B is 25% of mass of A then de-Broglie wavelength of B if wavelength of A is $$2\mathring A$$

A
$$4A^{\circ}$$

B
$$20A^{\circ}$$

C
$$6.0A^{\circ}$$

D
none

Solution
Question 10
1 / -0

A graph is plotted between uncertainty in position and inverse of uncertainty in wavelength for an electron. We get a straight line passing through the origin. Calculate voltage through which electron is accelerated with

A
150 V

B
75 V

C
37.5 V

D
300 V

Solution

$$Slope = \displaystyle\frac { \Delta x }{ { 1 }/{ \Delta \lambda } } =\Delta x\cdot \Delta \lambda $$
By Heisenberg principle $$\Delta x\cdot \Delta \lambda = \displaystyle\frac { { \lambda }^{ 2 } }{ 4\pi } $$
$$\displaystyle\frac { { \lambda }^{ 2 } }{ 4\pi } = \displaystyle\frac { 1 }{ \pi } \Rightarrow \lambda =2\mathring { A } $$
$$\lambda = \sqrt { \displaystyle\frac { 150 }{ V } } \Rightarrow V=37.5V$$

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Metal	$$W_0(eV)$$
Barium	2.5
Lithium	2.3
Tantalum	4.2
Tungsten	4.5

Dual Nature of Radiation and Matter Test - 65

Result

Dual Nature of Radiation and Matter Test - 65

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SHARING IS CARING

Question 1

Energy

Velocity

Momentum

Angular momentum

Solution

Question 2

Barium only

Barium or lithium

Lithium, tantalum or tungsten

Tungsten or tantalum

Solution

Question 3

Will be possible

Will not be possible

Data is incomplete

Depends upon the density of the surface

Solution

Question 4

Statement-1 is false, Statement -2 is true

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 correct explanation for statement-1

Statement-1 is true, Statement -2 is false

Solution

Question 5

$$1.4 \times 10^{19}$$

$$1.4 \times 10^{20}$$

$$2 \times 10^{19}$$

$$2 \times 10^{20}$$

Solution

Question 6

$$\dfrac{h}{m_0c}$$

$$\dfrac{m_0c}{h}$$

Zero

$$\infty $$

Solution

Question 7

$$400 nm$$

$$200 nm$$

$$4 \times 10 ^{-10} \mathring { A } $$

None of these

Solution

Question 8

1 A

1.23 A

2.46 A

None of these

Solution

Question 9

$$4A^{\circ}$$

$$20A^{\circ}$$

$$6.0A^{\circ}$$

none

Solution

Question 10

150 V

75 V

37.5 V

300 V

Solution

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