Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

What wavelength must electromagnetic radiation have if a photon in the beam has the same momentum as an electron moving with a speed $$1.1\times 10^5$$m/s(Planck's constant $$=6.6\times 10^{-34}$$J-s, rest mass of electron$$=9\times 10^{-31}$$kg?

A
$$\dfrac{2}{3}$$nm

B
$$\dfrac{20}{3}$$nm

C
$$\dfrac{4}{3}$$nm

D
$$\dfrac{40}{3}$$nm

E
$$\dfrac{3}{20}$$nm

Solution

Given,
$$m_e=$$mass of electron
$$=9\times 10^{-31}kg$$

$$v_e=1.1\times 10^{5}m/s$$

Momentum, $$p=m_ev_e$$

$$p=9\times 10^{-31}\times 1.1\times 10^5$$
$$=9.9\times 10^{-26}kg\ m/s$$

From de-Broglie waves
$$\lambda =\dfrac{h}{p}=\dfrac{6.6\times 10^{-34}}{9.9\times 10^{-26}}m$$

$$=\dfrac{2}{3}\times 10^{-8}m$$

$$=\dfrac{20}{3}\times 10^{-9}m=\dfrac{20}{3}nm$$
Question 2
1 / -0

To increase de-Broglie wavelength of an electron from $$0.5\times 10^{-10}$$m to $$10^{-10}$$m, its energy should be _______.

A
Increased to $$4$$ times

B
Halved

C
Doubled

D
Decreases to fourth part

Solution

$$\lambda =\dfrac{h}{p}=\dfrac{h}{\sqrt{2mK}}$$
$$\lambda \rightarrow 2\lambda$$
$$\Rightarrow K\rightarrow \dfrac{K}{4}$$.
Question 3
1 / -0

If the de Broglie wavelength of a proton is $$10^{-13} m$$, the electric potential through which it must have been accelerated is

A
$$4.07\times 10^{4}V$$

B
$$8.2\times 10^{4}V$$

C
$$8.2\times 10^{3}V$$

D
$$4.07\times 10^{5}V$$

Solution
Question 4
1 / -0

Five volts of stopping potential is needed for the photoelectrons emitted out of a surface of work function $$2.2\,eV$$ by the radiation of wavelength

A
$$1719\,\overset{\circ}{A}$$

B
$$3444\,\overset{\circ}{A}$$

C
$$861\,\overset{\circ}{A}$$

D
$$3000\,\overset{\circ}{A}$$

Solution

Einstein's equation for the photoelectric effect is $$eV=\dfrac {hc}{\lambda} - W $$

$$\dfrac {hc}{\lambda}=5\,eV+2.2\,eV=7.2\,eV$$

$$7.2=\dfrac{12375}{\lambda(in \overset{\circ}{A})}$$

or $$\lambda(in \overset{\circ}{A})=\dfrac{12375}{7.2}=1719\ \overset{\circ}{A}$$
Question 5
1 / -0

The voltage applied to an X-ray tube is $$18 \ kV$$. The maximum mass of photon emitted by the X-ray tube will be

A
$$2 \times 10^{-13} \ kg$$

B
$$3.2 \times 10^{-36} \ kg$$

C
$$3.2 \times 10^{-32} \ kg$$

D
$$9.1 \times 10^{-31} \ kg$$

Solution

Energy of photon is given by $$mc^2$$. Now, the maximum energy of photon is equal to the maximum energy of electrons $$=eV$$

Hence, $$ mc^2 = eV$$
$$\implies m= \dfrac{eV}{c^2} $$
$$ = \dfrac{1.6 \times 10^{-19} \times 18 \times 10^3}{(3 \times 10^8)^2} = 3.2 \times 10^{-32} \ kg$$
Question 6
1 / -0

A particle of mass $$10^{-31},kg$$ is moving with a velocity equal to $$10^5 \,ms^{-I}$$ . The wavelength of the particle is equal to

A
$$0$$

B
$$6.6\times10^{-8}\,m$$

C
$$0.66\,m$$

D
$$1.5\times10^{-7}\,m$$

Solution

As we know,
$$\lambda=\dfrac{h}{p}$$

$$\implies\lambda=\dfrac{6.6\times10^{-34}}{10^{-31}\times10^5}$$

$$\lambda=6.6\times10^{-8}\,m$$
Question 7
1 / -0

The maximum kinetic energy of the photoelectrons emitted from a surface is dependent on the :

A
Intensity of incident radiation

B
Potential of the collector electrode

C
Frequency of incident radiation

D
Angle of incident of radiation of the surface

Solution
Question 8
1 / -0

Einstein's photoelectric equation states that $$E_{k} = hv - \phi$$. In this equation $$E_{k}$$ refers to

A
Kinetic energy of all the emitted electrons

B
Mean kinetic energy of the emitted electrons

C
Maximum kinetic energy of the emitted electrons

D
Minimum kinetic energy of the emitted electrons

Solution
Question 9
1 / -0

The de-Broglie wavelength associated with with the particle of mass $$m$$ moving with velocity $$v$$ is

A
$$h/mv$$

B
$$mv/h$$

C
$$mh/v$$

D
$$m/vh$$

Solution

The de-Broglie wavelength associated with with the particle of mass $$m$$ moving with velocity $$v$$ is
$$\lambda =\dfrac hp =\dfrac{h}{mv}$$
Question 10
1 / -0

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of $$3\ km/s$$ will be

A
$$1\ A^o$$

B
$$0.66\ A^o$$

C
$$6.6\ A^o$$

D
$$66\ A^o$$

Solution

As we know,
$$\lambda =\dfrac {h}{mv_{rms}}\Rightarrow \lambda =\dfrac {6.6 \times 10^{-34}}{2\times 1.67 \times 10^{-27}\times 3\times 10^3}=0.66\ A^o$$

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Dual Nature of Radiation and Matter Test - 57

Result

Dual Nature of Radiation and Matter Test - 57

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SHARING IS CARING

Question 1

$$\dfrac{2}{3}$$nm

$$\dfrac{20}{3}$$nm

$$\dfrac{4}{3}$$nm

$$\dfrac{40}{3}$$nm

$$\dfrac{3}{20}$$nm

Solution

Question 2

Increased to $$4$$ times

Halved

Doubled

Decreases to fourth part

Solution

Question 3

$$4.07\times 10^{4}V$$

$$8.2\times 10^{4}V$$

$$8.2\times 10^{3}V$$

$$4.07\times 10^{5}V$$

Solution

Question 4

$$1719\,\overset{\circ}{A}$$

$$3444\,\overset{\circ}{A}$$

$$861\,\overset{\circ}{A}$$

$$3000\,\overset{\circ}{A}$$

Solution

Question 5

$$2 \times 10^{-13} \ kg$$

$$3.2 \times 10^{-36} \ kg$$

$$3.2 \times 10^{-32} \ kg$$

$$9.1 \times 10^{-31} \ kg$$

Solution

Question 6

$$0$$

$$6.6\times10^{-8}\,m$$

$$0.66\,m$$

$$1.5\times10^{-7}\,m$$

Solution

Question 7

Intensity of incident radiation

Potential of the collector electrode

Frequency of incident radiation

Angle of incident of radiation of the surface

Solution

Question 8

Kinetic energy of all the emitted electrons

Mean kinetic energy of the emitted electrons

Maximum kinetic energy of the emitted electrons

Minimum kinetic energy of the emitted electrons

Solution

Question 9

$$h/mv$$

$$mv/h$$

$$mh/v$$

$$m/vh$$

Solution

Question 10

$$1\ A^o$$

$$0.66\ A^o$$

$$6.6\ A^o$$

$$66\ A^o$$

Solution

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