Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 36

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Question 1
1 / -0

If the de-Broglie wavelength of a particle of mass $$m$$ is $$100$$ times its velocity then its value in terms of its mass $$(m)$$ and Planck's constant $$(h)$$ is:

A
$$\dfrac {1}{10}\sqrt {\dfrac {m}{h}}$$

B
$$10\sqrt {\dfrac {h}{m}}$$

C
$$\dfrac {1}{10}\sqrt {\dfrac {h}{m}}$$

D
$$10\sqrt {\dfrac {m}{h}}$$

Solution

$$\lambda ={\dfrac {h}{mv}}$$

According to given question $$ \lambda = 100v $$

$$100v={\dfrac {h}{mv}}$$

so $$v^2=\dfrac{h}{100m}$$

$$ \lambda = 100v $$

$$ \lambda= 10 \sqrt {\dfrac{h}{ m}}$$

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

Work function of a metal is $$5.2\times 10^{-18}\ J$$, then its threshold wavelength will be

A
$$736.7\mathring{A}$$

B
$$760.7\mathring{A}$$

C
$$301\mathring{A}$$

D
$$380.7\mathring{A}$$

Solution

Given - work function of metal $$W=5.2\times10^{-18}J$$
The work function of a metal is given by ,
$$W=\dfrac{hc}{\lambda_{0}}$$
or $$\lambda_{0}=\dfrac{hc}{W}$$
where $$\lambda_{0}=$$ threshold wavelength ,
$$h=6.6\times10^{-34}Js$$
$$c=3\times10^{8}m/s$$
Hence ,by $$\lambda_{0}=\dfrac{hc}{W}$$
$$\lambda_{0}=\dfrac{6.6\times10^{-34}\times3\times10^{8}}{5.2\times10^{-18}}=380.7A^{0}$$
Question 3
1 / -0

Photoelectrons are liberated by ultraviolet light of wavelength $$3000\mathring{A}$$ from a metallic surface for which the photoelectric threshold is $$4000\mathring{A}$$. The de-Broglie wavelength of electrons emitted with maximum kinetic energy is:

A
$$1.2 \ nm$$

B
$$3.215 \ nm$$

C
$$7.28$$ $$\mathring{A}$$

D
$$1.65$$ $$\mathring{A}$$

Solution

Kinetic energy= Quantum energy - Threshold energy

$$K.E= \dfrac{6.625\times 10^{-34}\times 3\times 10^{8}}{3000\times 10^{-8}}-\dfrac{6.625\times 10^{-34}\times 3\times 10^{8}}{4000\times 10^{-8}}=1.6565\times 10^{-19}$$J

$$\dfrac{1}{2}mv^{2}= 1.6565\times 10^{-19}$$

$$m^{2}\times v^{2}= 2\times 1.6565\times 10^{-19}\times 9.1\times 10^{-31}$$

$$mv= 5.49\times 10^{-25}$$

$$\lambda= \dfrac{h}{mv}$$

$$\lambda= \dfrac{6.625\times 10^{-34}}{5.49\times 10^{-25}}= 1.2\times 10^{-9}$$m

I.e 1.2 nm ($$1nm=10^{-9}m$$)

Hence option $$A$$ is correct.
Question 4
1 / -0

The work function of metals is in the range of $$2$$ eV to $$5$$ eV. Find which of the following wavelength of light cannot be used for photoelectric effect. (Consider, Planck constant $$=4\times10^{-15}$$eVs, velocity of light $$= 3\times 10^8m/s$$).

A
$$510nm$$

B
$$650nm$$

C
$$400nm$$

D
$$570nm$$

Solution

The minimum wavelength to cause photoelectric effect $$=\dfrac{hc}{E_{max}}$$
$$=\dfrac{4\times 10^{-15}\times 3\times 10^8}{5}m$$
$$=240nm$$
The maximum wavelength to cause photoelectric effect $$=\dfrac{hc}{E_{min}}$$
$$=\dfrac{4\times 10^{-15}\times 3\times 10^8}{2}m$$
$$=600nm$$
Hence light of wavelength $$650nm$$ is not acceptable.
Question 5
1 / -0

The work function of a metallic substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from the substance is approximately..

A
220 nm

B
310 nm

C
400 nm

D
540 nm

Solution

Work function of metallic substance $$\phi= 4.0 \ eV$$
For photoelectric effect to take place, minimum energy of the incident photon must be equal to the work function of metal surface i.e. $$E_{min} = \phi$$
$$\therefore$$ $$\dfrac{hc}{\lambda_{max}} = \phi$$
Or $$\lambda_{max} = \dfrac{hc}{\phi}$$
Or $$\lambda_{max} = \dfrac{1240}{\phi \ (in \ eV)} \ nm$$
$$\implies \ \lambda_{max} = \dfrac{1240}{4} = 310 \ nm$$
Question 6
1 / -0

A ruby laser produces radiations of wavelengths, 662.6 nm in pulse duration are 10$$^{-6}$$s. If the laser produces 0.39 J of energy per pulse, how many photons are produced in each pulse?

A
$$1.3 \times 10^9$$

B
$$1.3 \times 10^{18}$$

C
$$1.3 \times 10^{27}$$

D
$$3.9 \times 10^{18}$$

Solution

Total energy of pulse, $$E=n\dfrac{hc}{\lambda}$$
$$\implies n=\dfrac{E\lambda}{hc}$$ $$=\dfrac{0.39\times 662.6\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^{8}}$$ $$=1.3\times 10^{18}$$
Question 7
1 / -0

The stopping potential as function of frequency of incident radiation is plotted for two different photoelectric surfaces A and B. The graphs show that the work function of A is

A
greater than that of B

B
lesser than that of B

C
equal than that of B

D
no comparison

Solution

Hint: Use Einstein equation for photoelectric emission.

Explanation:-
Step 1: Concept used:
From Einstein’s equation ,
$$ eV_0 = h\nu – h\nu_0 $$
$$ V_0 = \dfrac{h\nu}{e} - \dfrac{h\nu_0}{e} $$
Step 2: Conclusion:

As work function $$ \phi =h \nu_0$$
$$\therefore \phi \propto \nu_0$$
From the graph, the threshold frequency of A $$(\nu_o)_A$$ is less than that of B $$(\nu_o)_B$$.
So, work function of A is less than B.
Hence option B is correct
Question 8
1 / -0

A particle of mass 1kg is moving with a velocity of 1m/s. The de-broglie wavelength associated with it will be

A
h

B
h/2

C
h/4

D
2h

Solution

Given : $$v = 1m/s$$ $$m = 1 \ kg$$
de-Broglie wavelength $$\lambda = \dfrac{h}{mv}$$
$$\therefore$$ $$\lambda = \dfrac{h}{1\times 1} = h$$
Question 9
1 / -0

The de-Broglie wavelength of a proton and alpha particle is same, the ratio of their velocities is :

A
1:2

B
2:1

C
1:4

D
4:1

Solution

de-Broglie wavelength, $$\lambda = \dfrac{h}{mv}$$

$$\implies \ v = \dfrac{h}{m\lambda}$$

We get $$\dfrac{v_{p}}{v_{\alpha}} = \dfrac{m_{\alpha}}{m_p}$$

Given : $$m_{\alpha} = 4m_p$$

$$\therefore$$ $$\dfrac{v_{p}}{v_{\alpha}} = \dfrac{4m_p}{m_p} = 4$$
Question 10
1 / -0

Electrons used in an electron microscope are accelerated by a voltage of $$25\ kV$$. If the voltage is increased by $$100\ kV$$ then the de-Broglie wavelength associated with electrons would

A
increase by $$2$$ times

B
decrease by $$2$$ times

C
increase by $$4$$ times

D
decrease by $$4$$ times

Solution

de_broglie wavelength $$\lambda = \dfrac{h}{P}$$
where momentum of the particle $$P = \sqrt{2mK} = \sqrt{2meV}$$
$$\implies \ \lambda = \dfrac{h}{\sqrt{2meV}}$$
We get $$\lambda\propto \dfrac{1}{\sqrt{V}}$$
Given : $$V_i = 25kV$$ and $$V_2 = 100kV$$
So, voltage is increased by a factor of $$4$$.
Thus wavelength gets reduced by a factor of $$2$$.

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Dual Nature of Radiation and Matter Test - 36

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Dual Nature of Radiation and Matter Test - 36

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Question 1

$$\dfrac {1}{10}\sqrt {\dfrac {m}{h}}$$

$$10\sqrt {\dfrac {h}{m}}$$

$$\dfrac {1}{10}\sqrt {\dfrac {h}{m}}$$

$$10\sqrt {\dfrac {m}{h}}$$

Solution

Question 2

$$736.7\mathring{A}$$

$$760.7\mathring{A}$$

$$301\mathring{A}$$

$$380.7\mathring{A}$$

Solution

Question 3

$$1.2 \ nm$$

$$3.215 \ nm$$

$$7.28$$ $$\mathring{A}$$

$$1.65$$ $$\mathring{A}$$

Solution

Question 4

$$510nm$$

$$650nm$$

$$400nm$$

$$570nm$$

Solution

Question 5

220 nm

310 nm

400 nm

540 nm

Solution

Question 6

$$1.3 \times 10^9$$

$$1.3 \times 10^{18}$$

$$1.3 \times 10^{27}$$

$$3.9 \times 10^{18}$$

Solution

Question 7

greater than that of B

lesser than that of B

equal than that of B

no comparison

Solution

Question 8

h

h/2

h/4

2h

Solution

Question 9

1:2

2:1

1:4

4:1

Solution

Question 10

increase by $$2$$ times

decrease by $$2$$ times

increase by $$4$$ times

decrease by $$4$$ times

Solution

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