Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 32

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Question 1

1 / -0

The work function for photoelectric effect

Is different for different metals

Is same for all metals

Depends upon the intensity of incident light

Depends upon the frequency of incident light

Solution

In solid-state physics, the work function (sometimes spelled workfunction) is the minimum thermodynamic work (i.e. energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface.
The work function is not a characteristic of a bulk material, but rather a property of the surface of the material (depending on crystal face and contamination), and varies for different metals.
The work function of some metals is given below as an eg,

Element	Work Function(eV)
Aluminum	4.08
Beryllium	5.0
Cadmium	4.07
Calcium	2.9
Carbon	4.81
Cesium	2.1
Cobalt	5.0
Copper	4.7
Gold	5.1
Iron	4.5
Lead	4.14
Magnesium	3.68
Mercury	4.5
Nickel	5.01
Niobium	4.3
Potassium	2.3
Platinum	6.35
Selenium	5.11
Silver	4.26-4.73*
Sodium	2.28
Uranium	3.6
Zinc	4.3

Question 2
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The average speed of a helium atom at $$25^oC$$ is $$1.23\times 10^3 ms^{-1}$$. What is the average wavelength of helium atom at $$25^oC$$?

A
$$1.35\times 10^{-31}$$

B
$$1.35\times 10^{-34}$$

C
$$1.35\times 10^{-30}$$

D
$$1.35\times 10^{-32}$$

Solution

According to de-Broglie's principle,
$$\lambda = h/mv$$ so
$$\lambda_{av}=\dfrac {h}{mu_{av}}=\dfrac {6.626\times 10^{-34}}{4\times 10^{-3}\times 1.23\times 10^3}$$
$$=1.35\times 10^{-34}$$
Question 3
1 / -0

If the kinetic energy of a free electron is doubled, its De Broglie wavelength changes by the factor:

A
$$\sqrt 2$$

B
$$\displaystyle\frac{1}{\sqrt 2}$$

C
$$2$$

D
$$1/2$$

Solution

As we know,
$$\quad \lambda =\frac { h }{ mv } =\frac { h }{ \sqrt { 2mE } } $$
so as KE doubles its De Broglie wavelength changes by
$$\lambda_2/\lambda_1 = 1/\sqrt { 2 }$$
Question 4
1 / -0

An electron is in an excited state in a hydrogen like atom. It has a total energy of $$-3.4\space eV$$. The kinetic energy of electron is $$'E\ '$$ and its de Broglie wavelength is $$'\lambda\ '$$

A
$$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-10}\space m$$

B
$$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-10}\space m$$

C
$$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-11}\space m$$

D
$$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-11}\space m$$

Solution

In any state,
$$KE = E, and PE = -2E$$, hence the total energy is $$-E.$$
Here $$E = 3.4eV.$$
Converting $$E$$ into $$Joules$$ and using the relation.

$$E = \dfrac{hc}{\lambda}$$

$$\lambda = \dfrac{hc}{E} = 6.6 * 10^{-10} m$$
Question 5
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An electron of mass 'm' and charge 'w' initially at rest gets accelerated by a constant electric field 'E'. The rate of change of de-Broglie wavelength of this electron at time 't' ignoring relativistic effects is

A
$$-\dfrac{h}{eEt^2}$$

B
$$-\dfrac{eht}{E}$$

C
$$-\dfrac{mh}{eEt^2}$$

D
$$\dfrac{h}{eE}$$

Solution

ginen, $$u=0$$
$$F = Ee$$
$$a =\cfrac{F}{m} = \cfrac{Ee}{m}$$
Newton's equation of motion
$$v=u+at$$
$$v= \cfrac{Eet}{m}$$
Debroglie Wavelength
$$\lambda = \cfrac{h}{mv} = \cfrac{h}{Eet}$$
Rate of change
$$\cfrac{d \lambda}{dt} = -\cfrac{h}{Eet^2}$$
Question 6
1 / -0

Which one of the following graphs represents the variation of maximum kinetic energy $$(E_k)$$ of the emitted electrons with frequency $$'\upsilon\ ' $$ in photoelectric effect correctly ?

A

B

C

D

Solution

The Einstein's photoelectric equation , $$h\nu=E_{k} +W$$
or $$E_k=h\nu-W$$
If we compare this equation $$y=mx+c$$ then slope $$m=h $$ and intercept $$c=-W$$
Question 7
1 / -0

Einstein's photoelectric equation is $$E_k=hv-\phi$$. In this equation $$E_k$$ refers to :

A
kinetic energy of all emitted electrons

B
mean kinetic energy of emitted electrons

C
maximum kinetic energy of emitted electrons

D
minimum kinetic energy of emitted electrons

Solution

Using Photoelectric effect

$${ E }_{ Incoming }=\phi +{ \left( KE \right) }_{ max }$$

Hence

$${ \left( KE \right) }_{ max }={ E }_{ Incoming }-\phi \quad $$

$$=hv-\phi $$

where $${ \left( KE \right) }_{ max }$$ is the maximum kinetic Energy of emitted electrons from the metal surface
Question 8
1 / -0

Which metal will be suitable for a photoelectric cell using light of wavelength $$4000 \mathring {A } $$ The work functions of sodium and copper are respectively 2eV and 4eV

A
Sodium

B
Copper

C
Both

D
None of these

Solution

Energy of incident radiation is given by:

$$E = \dfrac{hc}{\lambda} = \dfrac{3* 10^{8} * 6.64 * 10^{-34}}{4000 * 10^{-10}} = 4.98 * 10^{-19} = 3.1125 eV$$

Hence it can be used to excite any metal surface whose work function is less than $$3.1125 eV.$$
Question 9
1 / -0

A proton and $$\alpha$$-particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelength will be

A
1 : 1

B
1 : 2

C
2 :1

D
$$2\sqrt 2 : 1$$

Solution

De Broglie wavelength is given by:

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

Writing momentum as a a function of kinetic energy and mass

$$\lambda = \dfrac{h}{\sqrt{2Em}}$$

and Kinetic energy = potential through which it is accelerated times the charge on it

$$\lambda = \dfrac{h}{\sqrt{2Vqm}}$$

$$\dfrac{\lambda_{proton}}{\lambda_{alpha}} = \sqrt{\dfrac{q_{alpha}m_{alpha}}{q_{proton}m_{proton}}}$$

$$q_{alpha} = 4e$$
$$q_{proton} = e$$
$$m_{alpha} = 4m_{proton}$$

Hence, $$\dfrac{\lambda_{proton}}{\lambda_{alpha}} = \sqrt{\dfrac{8}{1}} = 2\sqrt{2}$$
Question 10
1 / -0

If the kinetic energy of a moving particle is E, then the de-Broglie wavelength is

A
$$\lambda =h \sqrt{2 m E}$$

B
$$\lambda =\sqrt{\dfrac{2 m E}{h}}$$

C
$$\lambda =\dfrac{h}{\sqrt{2 m E}}$$

D
$$\lambda =\dfrac{hE}{\sqrt{2 m E}}$$

Solution

$$ E=\dfrac{1}{2}mv^2 \,\,or\,\,$$

$$ mv= \sqrt{2mE} \,\, or\,\,$$

$$ \lambda =\dfrac{h}{mv}=\dfrac{h}{\sqrt{2mE}}$$

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Dual Nature of Radiation and Matter Test - 32

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Dual Nature of Radiation and Matter Test - 32

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Question 1

Is different for different metals

Is same for all metals

Depends upon the intensity of incident light

Depends upon the frequency of incident light

Solution

Question 2

$$1.35\times 10^{-31}$$

$$1.35\times 10^{-34}$$

$$1.35\times 10^{-30}$$

$$1.35\times 10^{-32}$$

Solution

Question 3

$$\sqrt 2$$

$$\displaystyle\frac{1}{\sqrt 2}$$

$$2$$

$$1/2$$

Solution

Question 4

$$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-10}\space m$$

$$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-10}\space m$$

$$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-11}\space m$$

$$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-11}\space m$$

Solution

Question 5

$$-\dfrac{h}{eEt^2}$$

$$-\dfrac{eht}{E}$$

$$-\dfrac{mh}{eEt^2}$$

$$\dfrac{h}{eE}$$

Solution

Question 6

Solution

Question 7

kinetic energy of all emitted electrons

mean kinetic energy of emitted electrons

maximum kinetic energy of emitted electrons

minimum kinetic energy of emitted electrons

Solution

Question 8

Sodium

Copper

Both

None of these

Solution

Question 9

1 : 1

1 : 2

2 :1

$$2\sqrt 2 : 1$$

Solution

Question 10

$$\lambda =h \sqrt{2 m E}$$

$$\lambda =\sqrt{\dfrac{2 m E}{h}}$$

$$\lambda =\dfrac{h}{\sqrt{2 m E}}$$

$$\lambda =\dfrac{hE}{\sqrt{2 m E}}$$

Solution

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