Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The green graph and the purple graph pictured below represent the photoelectric effect for two different metals (G and P).
If electromagnetic radiation of frequency $$11\times 10^{14}Hz$$ strikes both metal surfaces, which of the following statements must be true?

A
The radiation will have to be more intense to knock electrons off the surface of metal P than metal G

B
No electrons will be knocked off the surface of either metal as a result of the radiation

C
Electrons will be knocked off both metal surfaces, but they will be knocked off at a greater rate from the surface of metal G

D
Electrons will be knocked off the surface of metal G by the radiation, but will not be knocked off the surface of metal P

E
Electrons will be knocked off both metal surfaces by the radiation, but the most energetic electrons knocked off the surface of metal G will have more kinetic energy than the most energetic electrons knocked off the surface of metal P

Solution

As the frequency of the EM radiation used is greater than the threshold frequency of both the metal P and G, thus electrons will be knocked off both metal surfaces by the radiation.
Maximum kinetic energy of the photoelectrons $$K.E_{max} = h\nu - \phi$$
Also from graph, we can state that work function $$(\phi)$$ of metal P is greater than that of G i.e $$\phi_P>\phi_G$$
$$\implies$$ $$K.E_G{_{max}} > K.E_{P_{max}}$$
Hence option E is correct.
Question 2
1 / -0

Calculate the de Brogile wavelength of a photon whose linear momentum has a magnitude of $$3.3 \times 10^{-23}kgm/s$$.

A
0.0002 nm

B
0.002 nm

C
0.02 nm

D
0.2 nm

E
2 nm

Solution

The de Brogile wavelength of a photon is $$\lambda=\dfrac{h}{p}=\dfrac{6.6\times 10^{-34}}{3.3\times 10^{-23}}=2\times 10^{-11}=0.02\times 10^{-9} m=0.02 nm$$
Question 3
1 / -0

The de Broglie wavelength of a tennis ball of mass $$60$$ g moving with a velocity of $$10$$ m/sec is approximately :

A
$$10^{-33}$$ m

B
$$10^{-31}$$ m

C
$$10^{-16}$$ m

D
$$10^{-25}$$ m

Solution

De-Broglie wavelenght is given by:
$$\lambda=\dfrac{h}{mv}=\dfrac{6.62\times10^{-34}}{60\times10^{-3}\times10} \approx10^{-33}$$ m
Question 4
1 / -0

The number of De-Broglie wavelengths contained in the second Bohr orbit of Hydrogen atom is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

De-broglie wavelength of a particle is given as
$$\lambda=\dfrac{h}{mv}$$
According to the Bohr's postulate,
$$mvr=\dfrac{nh}{2\pi}$$
$$\implies 2\pi r=n\dfrac{h}{mv}=n\lambda$$
Hence there are $$n$$ De-Broglie wavelengths in $$n^{th}$$ Bohr orbit.
Hence correct answer is option B.
Question 5
1 / -0

What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 Volt?

A
12.27 $$\mathring{A}$$

B
1.227 $$\mathring{A}$$

C
0.1227 $$\mathring{A}$$

D
0.001227 $$\mathring{A}$$

Solution

Given : $$V =100$$ volts
de-Broglie wavelength of electron $$\lambda = \dfrac{h}{\sqrt{2meV}}$$
$$\therefore$$ $$\lambda = \dfrac{6.6\times 10^{-34}}{\sqrt{2(9.1\times 10^{-31})(1.6\times 10^{-19})(100)}}$$

Or $$\lambda = \dfrac{6.6\times 10^{-34}}{5.4\times 10^{-10}}$$ m
$$\implies $$ $$\lambda = 1.227\times 10^{-10} m = 1.227$$ $$\mathring{A}$$
Question 6
1 / -0

Determine de-Broglie wavelength of an electron having kinetic energy of $$1.6 \times 10^{-6}$$ erg.

A
$$12.42A^o$$

B
$$12.42 cm$$

C
$$12.42 m$$

D
none of these

Solution

Debroglie wavelength is given by $$ \lambda = \dfrac {h}{mv}$$
When KE is given , then $$ \lambda = \dfrac {h}{{(2m KE)}^{1/2}}$$

If m= mass of electron in kg , KE in Joules , then answer comes out in Angstrom.
Mass of electron =$$ 9.1× {10}^{-31}$$ kg
For conversion use :: 1 Joules = $${10}^{7}$$ erg.
Question 7
1 / -0

Find the de-Broglie wavelength of an electron with kinetic energy of $$120\ eV$$.

A
$$95\ pm$$

B
$$102\ pm$$

C
$$112\ pm$$

D
$$124\ pm$$

Solution

Given : $$K = 120eV$$

Mass of electron $$m = 9.1\times 10^{-31}$$ kg

de-Broglie wavelength $$\lambda= \dfrac{h}{\sqrt{2mK}}$$

$$\therefore$$ $$\lambda= \dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 120 \times 1.6\times 10^{-19}}}$$

Or $$\lambda = \dfrac{6.6\times 10^{-34}}{59.1\times 10^{-25}} = 0.112\times 10^{-9}$$ $$m = 112$$ $$pm$$
Question 8
1 / -0

If the momentum of an electron is increased by $${ P }_{ m }$$ then the de Broglie wavelength associated with it changes by $$0.5$$%. Then the initial momentum of the electron is

A
$$100{ P }_{ m }$$

B
$$\dfrac { { P }_{ m } }{ 100 } $$

C
$$200{ P }_{ m }$$

D
$$\dfrac { { P }_{ m } }{ 200 } $$

Solution

The De broglie wavelength $$\lambda$$ and momentum $$p$$ of a particle are related by ,
$$p=\dfrac{h}{\lambda}$$
as $$p\propto 1/\lambda$$ , therefore , with increase in momentum , wavelength will decrease .
Initially , $$p=\dfrac{h}{\lambda}$$ ................eq1
Now the wavelength is decreased by 0.5% ,
$$\lambda'=\lambda-\lambda\times0.5/100=199\lambda/200$$
hence , increased momentum ,
$$p+p_{m}=\dfrac{h}{\lambda'}$$
or $$p+p_{m}=\dfrac{200h}{199\lambda}$$ ............eq2
dividing eq2 by eq1 , we get
$$\dfrac{p+p_{m}}{p}=\dfrac{200}{199}$$
or $$199p+200p_{m}=200p$$
or $$p=200p_{m}$$
Question 9
1 / -0

If a surface has a work function $$4.0 eV$$, what is the maximum velocity of electrons liberated from the surface when it is irradiated with ultraviolet radiation of wavelength $$0.2\mu m$$?

A
$$4.4\times { 10 }^{ 5 }{ m }/{ s }$$

B
$$8.8\times { 10 }^{ 7 }{ m }/{ s }$$

C
$$8.8\times { 10 }^{ 5 }{ m }/{ s }$$

D
$$4.4\times { 10 }^{ 7 }{ m }/{ s }$$

Solution

Given : $$\phi = 4eV$$ $$\lambda = 0.2\mu m = 200$$ $$nm$$
Using $$K.E_{max} = \dfrac{hc}{\lambda} -\phi$$ where $$K.E_{max} = \dfrac{1}{2}mv^2$$ and $$hc =1240$$ $$eV$$
$$\therefore$$ $$K.E_{max} = \dfrac{1240}{200} -4 = 2.2 eV = 2.2\times 1.6\times 10^{-19} = 3.53\times 10^{-19}$$ $$J$$
Maximum velocity $$v =\sqrt{\dfrac{2K.E_{max}}{m}}$$
$$\therefore$$ $$v = \sqrt{\dfrac{2\times 3.52\times 10^{-19}}{9.1\times 10^{-31}}} = 8.8\times 10^5$$ $$m/s$$
Question 10
1 / -0

The de Broglie wavelength of an electron (mass = $$1 \times 10^{-30}$$ kg, charge = $$1.6 \times 10^{-1} C$$) with a kinetic energy of 200 eV is (Planck's constant = $$6.6 \times 10^{-34} J s$$)

A
$$9.60 \times 10^{-11} m$$

B
$$8.25 \times 10^{-11} m$$

C
$$6.25 \times 10^{-11} m$$

D
$$5.00 \times 10^{-11} m$$

Solution

Given : $$m=1\times 10^{-30}$$ kg $$e =1.6\times 10^{-19} C$$ $$h =6.6\times 10^{-34} Js$$
Kinetic energy $$K =200eV$$
de Broglie wavelength $$\lambda = \dfrac{h}{\sqrt{2mK}} =\dfrac{h}{\sqrt{2m (200) e}}$$
$$\therefore$$ $$\lambda = \dfrac{6.6\times 10^{-34}}{\sqrt{2(10^{-30}) (200) (1.6\times 10^{-19})}}$$

OR $$\lambda = \dfrac{6.6\times 10^{-34}}{8\times 10^{-24}} $$ $$\implies$$ $$\lambda = 8.25\times 10^{-11} m$$

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Dual Nature of Radiation and Matter Test - 35

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Dual Nature of Radiation and Matter Test - 35

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Question 1

The radiation will have to be more intense to knock electrons off the surface of metal P than metal G

No electrons will be knocked off the surface of either metal as a result of the radiation

Electrons will be knocked off both metal surfaces, but they will be knocked off at a greater rate from the surface of metal G

Electrons will be knocked off the surface of metal G by the radiation, but will not be knocked off the surface of metal P

Electrons will be knocked off both metal surfaces by the radiation, but the most energetic electrons knocked off the surface of metal G will have more kinetic energy than the most energetic electrons knocked off the surface of metal P

Solution

Question 2

0.0002 nm

0.002 nm

0.02 nm

0.2 nm

2 nm

Solution

Question 3

$$10^{-33}$$ m

$$10^{-31}$$ m

$$10^{-16}$$ m

$$10^{-25}$$ m

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 5

12.27 $$\mathring{A}$$

1.227 $$\mathring{A}$$

0.1227 $$\mathring{A}$$

0.001227 $$\mathring{A}$$

Solution

Question 6

$$12.42A^o$$

$$12.42 cm$$

$$12.42 m$$

none of these

Solution

Question 7

$$95\ pm$$

$$102\ pm$$

$$112\ pm$$

$$124\ pm$$

Solution

Question 8

$$100{ P }_{ m }$$

$$\dfrac { { P }_{ m } }{ 100 } $$

$$200{ P }_{ m }$$

$$\dfrac { { P }_{ m } }{ 200 } $$

Solution

Question 9

$$4.4\times { 10 }^{ 5 }{ m }/{ s }$$

$$8.8\times { 10 }^{ 7 }{ m }/{ s }$$

$$8.8\times { 10 }^{ 5 }{ m }/{ s }$$

$$4.4\times { 10 }^{ 7 }{ m }/{ s }$$

Solution

Question 10

$$9.60 \times 10^{-11} m$$

$$8.25 \times 10^{-11} m$$

$$6.25 \times 10^{-11} m$$

$$5.00 \times 10^{-11} m$$

Solution

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