# KPK Board 12th class Physics Ch 18 Dawn of Modern Physics short questions answers

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Imagine a world in which a C = 50m/s. How would everyday events appear to us?

Decreasing the speed of light increases the relativistic factor 1 while increasing the speed of light decreasing the relativistic factor. Therefore if we imagine the speed of light C = 50m/se, then according to Lorentz transformation the value of V must be less than C. There would more effect on the mass m, length L of an object moving with velocity V. Moreover there would be more effect on time dilation, as the range between v and c has been decreased.

1. The relativistic mass is

m=mo/1-V2/C2

When C= V, then m becomes infinite.

1. Similar the length contraction is

L=Lo/

iii. The time dialation is

t=to/

1. Similarly the effect on Einstein mass energy relation is also greater. Because the change in mass Am will be greater when it moves with velocity V.

E = (Δm) C2

Thus when C = 50/mse, then the effect on every day events will be greater as compare to the present time.

Both Zarak and Samina are twenty years old. Zarak leaves earth in a space craft moving at 0.8C while Samina remains on the earth. Zarak returns from a trip to star 30 light years from the earth. Which one will be of greater age Explain.

This is a problem of twin paradox. Samina will be of greater age. According to special theory of time dilation, the moving clock looses compared to a stationary clock. In other words a moving clock runs slo compared to a stationary clock. The present age of each one is 20 years.  The ground point of wie, the time taken by Zarak coving 30 light years and return back to the earth is.

T=2S/V =2xl yearxCx30/0.8C=75Yea

The age of Samina when Zarak arrives the earth is.

Ts=Tz/1-V2/C2

Tz=TS1-V2/C2=Ts1-(0.8)2/C2

Samina age = Ts = 20+75 = 95 years.                                                                                                                     From the spaceship point of view the length of the journey is contracted which Zarak covers in less time. So, according to the theory of time dilation Zarak age is

Tz =95 0.36=95×0.6=75Year

Thus after trip Zarak will be of 57 years and Samina of 95 years old.

Which has more energy a photon of ultraviolet radiation or a photon of yellow light.

The photon of ultraviolet light has more energy than the photon of yellow light. This is because the wave length of ultraviolet is of the order of 3840Ao while that of yellow light is of the order of 5560Ao. Thus to calculate the energy of each photon we have

Speed of light =C = 3×10 m/se

Wavelength of ultraviolet light = 2x = 3840Ao = 3840×10-m

Wavelength of yellow light = 2y, = 5560A°=5560×10-10m

The energy of ultraviolet light photon is:

Energy of (Photon)u = hfu=h(c/2u)6.63×10-34x3x108/3840×10-10

Eu = 5.18×10-19J

The energy of yellow light Photon is given by

Energy of violet light photon = hf, = h(C/2y)=6.63×10-34x3x108/5560×10-10

Ey = 3.577 10-19

Thus from eq. (1) and (2) it is clear that ultraviolet photon has greater energy.

Some stars are observed to be radish and some are blue. Which stars ha higher surface temperature? Explain.

Those stars which are observed blue, they have higher surface temperature, because, blue light has wavelength of 4861A and red light wave length is 6563Ao. Now according to Wein’s displacement law, the wavelength of the radiation emitted is inversely proportional to the absolute temperature of the radiator wavelength lun of the maximum intensity is given by:

2m =con tanx1/T

T=con tan/2m

The value of constant = 0.0029m.kº

Wavelength of blue light=2b = 4861Ao = 4861×10-10m

Wave length of red light=2r, = 6563A° = 6563×10-10 m

Temperature of Blue stars = Tb = cons tan t/2b

T=0.0029/4861×10-10=5.967×103=5967K

Temperature of blue stars =Tr=Cons tan t/2r

Tr = 4.4187×103 = 4418.7k-

Thus the surface temperature of blue stars is higher than the radish stars. This is why we use blue light for welding rather than red light.

An electron and proton are accelerated from rest through the same potential differences. Which particle has the longer wavelength? Explain.

When an electron and proton are accelerated through the same potential difference, then electron will have longer wave length. Because mass of electron is very small as compare to proton. The de Broglie’s wavelength is

2=h/mV=h/P

But k.E=P2/2m

P=

Putting this value in equa (1)

2=h/

Now when acharge particle q is accelerated through P.D say V., its K.E is-

K.E = qVo

Putting this value in equ (3) we get

2=h2m qVo

When q and V. are kept constant then, 2 is inversely proportional to the square root of the mass of charge particle.

m2 1/                                                                                                                                                                              ear that the wavelength of electron is longer than proton as its mass, Thus it is clear that the wave is 1840 times less than proton-

All objects radiate energy. Explain why then we are not able to see objects in a dark room?

The objects in a room are in thermal equilibrium. Each object in a room radiates ad absorbs energy from the surroundings in the form of infrared light. But human eye is sensitive to visible light and can see those objects which radiate energy in the visible range of electromagnetic radiation spectra. Therefore, objects in a dark room radiate infrared radiations which are not visible to eye. The device snooper scope is used to see the warm object cumming infrared radion in dark. The snooper scope is used by initialy dark light to see objects.

If the photo electric effect is observed for one metal, can you conclude that the effect will also be observed for another metal under the same conditions?

No when photo electric effect is observed for one metal, under the influence of certain radiation, cannot be generalized for another metal. Because  the ejection of electrons from the metal surface depends upon its work function. When radiations of wavelength 2. Are incident on a metal surface of work function 6, the photoelectric effect will take place only if the energy of incident photon is equal or greater than the work function.

hf = 0

hf/c=

h/2=/c

2=ch/

21

Since different metals have different work function, so they will require radiations of different wavelengths, in order to eject electrons from their surface.

Explain why it is impossible for a particle with rest mass to move faster the speed of light?

It is not possible for a particle having rest mass to move with a speed equal to – greater than the speed of light. Because according to special theory of relativity the mass of the particle becomes infinite when V = C, it requires large force.

m = mo/ = mo/ = me/ = me/o =

F = ma =  X a =

As no material object can have infinite mass, so the speed of a particle can never be equal to C or greater than speed of light C.

Use photon model to explain why the ultraviolet is harmful to your skin while visible light is not?

The ultraviolet is short wavelength radiation ranging from 0.4um to about 1.nm, while visible light has long wavelength such as 7.0×10 m. Thus ultraviolet contain high frequency or energy photons as compare to ordinary visible light. The energy of photon is.

E=hf

Since f= V/2=C/2

So

E =Ch/2

Since 2e <22l thus E, will be greater than EL.

1. Harmful: Now harmful effects arise due to the absorbtion of radiation in the tissues, causing disorder in the normal function of the tissues. Visible ordinary light contains low energy photons, which have low penetrating power and safe.
2. Ultraviolet has high energy photons, produces vitamins in the skin and causes sun- tan. But an overdose can be harmful, especially causes eye and skin disease including skin cancer.
Explain why the annihilation of an electron and positron creates a pair of photons rather than a single photon?

To satisfy the law, the laws of conservation of energy and momentum, the two photons are in opposite direction when annihilation of an electron and positron take emitted in opposite place.

Explanation: Electron is a particle and positron is its anti-particle. When a particle combines with its anti-particle, pair annihilation takes place. There is a version of rest mass energy to electromagnetic energy and confirms the result.

E = mc2

1. According to the law of conservation of energy.

mo C + mo C2 = hf +hf

2moC2 = 2hfmo C2 = hf                                                                                                                                                                                                      hf = moC =9.11×10-31(3 x108)=0.51MeV                                                                                                                     2. According to the law of conservation of momentum

0+0 = hf/C + [hf/C]

Hf/C = hf/C

Thus to satisfy the laws of conservation of energy and momentum the two pho are emitted in opposite direction.

When a particle's K.E increases, what happens to its de Broglie wavelength

When K.E of a particle increases, its de Broglie wave length decreases. The de Broglie wavelength of a particle of mass m, moving with a velocity V, is given by.

2=h/mv =h/P

Where h is a Plank’s constant. But we know that the K.E and momentum Pare related as.

KE =P / 2 m

P=2m(K.E)

Putting this value in equ(1) we get

2=h/2 m (K.E)

21/(K.E)

Explain why we can experimentally, observe the wave like properties electrons but not of billiard ball?

The basic characteristics of a wave are its wavelength λ and frequency f.

The de Broglie’s wavelength 2 associated with a particle of mass m moving with a velocity V is given by

λ =h/p

λ=h/mV

Thus it is clear that the wavelength of a particle wave depends upon its mess in addition to its velocity.

1. Electron is a subatomic particle and its mass is very small. When it moves its wavelength is of the order of 1.65 x 10-10m which is of the order of the x-rays Wavelength and can be detected and measured experimentally by electron microscope.
2. The mass of billiard is very very large as compare to an electron. When it moves the wavelength associated with it is so small that it cannot be measured and detected even by an ideal instrument. Thus electron possess wave like properties but billiard ball does not.
Does a light bulb at a temperature of 2500 K produce as white light as the sun at 6000 K? Explain.

No, a light bulb at a temperature of 2500 K does not produce as white light as the sun at 6000 K. Because the visibility of light depends upon the frequency of light. But brightness of light depends upon the number of photons or intensity of light.

Explanation: From Boltzman-Stefan’s law, we know that the energy E radiated per second per unit area is directly proportional to the 4th power of the absolute temperature of the hot body.

Ε α Τ4

E = ƍT4

Where  ƍ is a constant. Using this formula, the energy distribution spectrum (curves) for each temperature are shown. As the temperature rises the intensity of the emitted light increases. That is why; the sun will have more bright light as compared to a light bulb 2500 K.

A beam of red light and a beam of blue light have exactly the same energy. Which light contains the greater number of photons?

A beam of red light contains the greater number of photons as compared to a beam of blue light. Because the wavelength of red light is greater than that of blue light, OR, the frequency of red light is smaller than the blue light. Moreover, the energy of a photon depends upon its frequency, and according to Einstein Quantum theory, the energy of a photon is.

E = hf

The energy of n number of photons is.

E = nhf

Since,

f = c/λ

therefore,

E = nch/λ

n = λE/ch

n directly proportional to λ.

Thus the number of photons is directly proportional to wavelength 2, so red light contains greater number of photons. In a Compton scattering experiment, an electron is accelerated straight ahead in the direction of the incident X-ray photon. Which way does the scattering photon move?

There are two possibilities, depending upon the energy of X-rays photon. a. In case of soft X-rays: If the incident photon is that of soft X-rays (low energy photons), then the relative mass of photon h/ac is less than the rest mass of electron m.. So the photon will bounce backward along its original path. Hence,

0=180°. Now the change in wavelength is given by.

λ’ – λ = h/mc (1-cos180°)

Δλ = h/mc [1-(-1)]

Δλ = 2h/mc

1. In case of hard X-rays (high energy): If the incident photon is that of hard x. rays then the relative mass of photon h/ac is greater than the rest mass of electron m. So after scattering the photon will move forward in its original direction and θ=0°. According to eq I the wave length shift is.

λ’ – λ = h/mc (1-cos0°) = h/mc(1-1)

Δλ = 0

Thus the wavelength shift is zero. The type of scattering in which the wavelength shift is zero is called Coherent scattering.

Why must the rest mass of photon be zero?

The rest mass of photon is zero, because we know that a photon is a packet of energy and moves with a speed of light C. If the rest mass m, of a photon is not zero, then according to Einstein’s special theory of relativity, the relative mass of photon “m” becomes infinite.

m = m̥/√1-v² /c²

m = m̥/√1-c² /c²

m = m̥/0 = 0

The energy possessed by photon also becomes infinite.i.e,

E = mc2

E = XC2 =

Since a photon does not possess infinite mass and energy, therefore, it is concluded that the rest mass m, of the photon must be zero.

What happens to total radiation from a black body if its absolute temperature is doubled?

Ans: By doubling the absolute temperature of a black body, its radiation will increase by 16 times. From Stefan’s-Boltzmen’s law, we know that the amount of energy of all wavelengths radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature T.

E T4

E T4

Now doubling the temperature is bounded, then T’ = 2T so,

E’ T)4

E’ T4 = E’

Thus by doubling the absolute temperature of a black body radiator, the radiant energy will increase by 16 times.

Why don't we observe Compton effect with visible light?

We do not observe Compton Effect with visible light. Because visible light photon has the energy comparable to the binding energy of the electron. It loses all of its energy in a single interaction with an electron in an atom and is disappeared that causes photoelectric effect.

For Compton Effect, the incident photon must have high energy and frequency. Compton Effect is observed with X-rays and y rays photons. These photons have energy which is thousand times greater than the visible light photons. During Compton Effect an X-ray photon ejects an electron from the atom. Even after this interaction the photon has sufficient energy, so that it retains its identity and is scattered with less energy and momentum. The wave length shift in Compton Effect is given by.

λ’ – λ = h/mc (1-cos0°)

Δλ = h/mc (1-cos0°)

If the following particles all have the same K.E, which has the shortest wavelength? Electron, alpha particle, neutron and proton?

Alpha particle will have the shortest wavelength, because its mass is very large as compared to the other particles. The de Broglie wave length associated with a small particle, such as electron, neutron and proton or alpha particle of mass m, moving with velocity V is given by.

λ = h/mV

λ = h/p

Now we know that the kinetic energy E of a particle is given by.

K.E = E = p2/2m

p2 = 2mE

P = √2mE

Putting this value in eq above we get.

λ = h/√2mE

λ = h/√2E x 1/√m

Since E is the same for all particles therefore, = Constant.

VE

λ = Constant x 1/√m

λ x 1/√m

Thus, the wavelength associated with a particle is inversely proportional to the square root of its mass. Since the mass of alpha particle is greater than is mass of proton or electron or neutron, so it is associated with shortest wavelength.

If an electron and a proton have the same de Broglie wavelength, which particle has greater speed?

When an electron and proton have the same wavelength then the electron has the greater speed on account of its smaller mass. The wavelength of electron is.

λ = h/meVe

Similarly the wavelength of proton is.

λ = h/mpVp

Comparing equs(1) and (2) we get

h/meVe = h/mpVp

meVe = mpVp

Ve/Vp = mp/me

Thus from equs (3) its clear that velocities are inversely proportional to the respective masses. Hence the electron has the greater speed than proton, as its mass is less than the mass of proton.

Use photon model to explain why the ultraviolet is harmful to your skin while visible light is not?

The ultraviolet is short wavelength radiation ranging from 0.4um to about 1.nm, while visible light has long wavelength such as 7.0×10 m. Thus ultraviolet contain high frequency or energy photons as compare to ordinary visible light. The energy of photon is.

E=hf

Since f= V/2=C/2

So

E =Ch/2

1. Harmful: Now harmful effects arise due to the absorbtion of radiation in the tissues, causing disorder in the normal function of the tissues. Visible ordinary light contains low energy photons, which have low penetrating power and safe.
2. Ultraviolet has high energy photons, produces vitamins in the skin and causes sun- tan. But an overdose can be harmful, especially causes eye and skin disease including skin cancer.
An incandescent light bulb is connected to a dimmer switch when the bulb operates at full power, it appears white, but as it is dimmed it looks more and more red. Explain?

Dimmer switch is a resistor introduced in series with incandescent light bulb to reduce the current in it. When the dimmer switch is turned ON! more resistance is offered and hence fewer current passes through the bulb. As current in the bulb reduces, so does the temperature and filament comes to a relatively lower temperature. With lower temperature, A max according to Wien’s displacement law (λmax T 0.2898 x 10-2 m ), shifts to larger values. On the other hand, when bulb operates at full power, maximum current flows only through the bulb, due to which bulb is at higher temperature and lower value of λmax.

Hence by introducing the dimmer switch, wavelength shifts from smaller to larger values leading to more and more reddish appearance of bulb. Dimmed bulbs waste most of the power as invisible infra-red light (heat) across the resistor.

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