# 12th Class Physics Chapter 8 DAWAN AND MODREN PHYSICS Short Question Answers

## 12th Class Physics Chapter 8 DAWAN AND MODREN PHYSICS Short Question Answers Below

- Yes, there would be an increase in the mass of compressed spring. However, this increase in mass is slightly greater than the mass of uncompressed spring due to condensation of energy of compressed spring into mass (i. e work done in compressing the spring has changed into energy which increase the mass of the spring) However, this increase in mass is neglibly small.

**Example:-**

Suppose a spring is compressed and energy stored in it is 10 j. Then by using the relation

ΔE = Δmc^{2}

or Δm = ΔE/c_{2}= 10/ (3*10^{8})^{2 }= 10/ 9*10^{16}

or Δm = 1.1 *10^{-15}kg

This increase in mass is negligible as compared to the original mass of spring.

E = T

^{4 }

When the temperature is increased to double of its value, then

E

^{/ }= σ (2T)

^{4}

Or E

^{/}= σ * 16 T

^{4}

E

^{/}= 16σ T

^{4}= 16E

Thus, the total radiation energy E

^{/}will increase 16 times.

E = hf

Let n be the number of photons, then the energy becomes

E = nh f …………………….(i)

But c = fλ or f = c/λ

Putting the value of f in equ (1) we have

E = nhc/λ

or n = Eλ/hc…………………………(2)

As a red and blue light have the same energies, so E, h and c have constant values. Thus

n = constant * λ

or nλ …………………………(3)

The relation (3) shows that greater wavelength will have a large number of photons.

But wavelength of red light is greater than that of blue light i.e . λ

_{red}>

λ

_{blue}, so beam of red light will contain a greater number of photons.

(a). As we know that energy of the photons given by

E = hf

But f= c/λ

E = hc/λ………………………..(1) (c =fλ)

But h and c have constant value. Therefore

E = constant * 1/λ

or E 1/λ

This relations shows, smaller the wavelength, larger the energy. As the wavelength of photon of blue light is shorter among red and green light photon, so

(a). Photon of blue light has maximum energy.

(b). As momentum of photon is given by

P = h/λ

But h is constant, therefore

P = cont * 1/λ

or p 1/λ

According to this relation, wavelength of blue light is shorter among red and green light photons. So

(b). Photon of blue light has maximum momentum.

- Energy of photon A = E
_{A}= hf_{A}

Energy of photon B = E_{B}= ½ hf_{A }(E=hf)_{ }Or E_{B}= ½ E_{A}…………………………………(1)

As we know that momentum of photon is given by

p = h/λ = hf /c = E/c (c = fλ)

or λ = c/f

Now

Momentum of photon A = p_{A} = E_{A}/c………………….(2)

Momentum of photon B = p_{B} = E_{B}/c

But from equ (1) we can write it

P_{B} = E_{B}/c = __ 1/2E _{A ……………………………………………….. }__(3)

c

Dividing equ (2) by equ (3) we get

P

_{A}/P

_{B}= E

_{A}/c * c/1/2E

_{A}= 1/1/2 = 2

or p √ p

_{B}= 2

Hence, momentum of photon A is also two times greater than the momentum of photon B.

λ = h/ mv

or v = h/mλ …………………………………(1)

As de Broglie wavelength λ for both the particles Is same end ‘h’ is also constant, so equ (1) can be written as

v = constant * 1/ m

or v 1/m

It shows that greater the mass of the particle, smaller will be its velocity. As mass of proton is greater than mass of the electron, so speed of electron would be greater than the speed of proton.

λ = h/ mv

Due to large mass of cricket ball and small speed. The value of wavelength ‘λ’ associated with a moving cricket ball is extremely small. So it cannot show the wave – propertmy and no diffraction can be produced. Hence, we cannot notice the de Broglie wave length for pitched cricket ball. In other words, wave length of wave associated with ball cannot be detected.

The minimum energy required to break the free electron from the metal surface is called the work function ‘φ’. According to the definition of work function, copper has greater work function than sodium.

Threshold Frequency:-

The relation between the work function and threshold frequency is given by

φ = hf_{o
}The relation shows that greater the work function, greater will be the threshold frequency ‘f_{o}’. Hence copper will have higher threshold frequency.

_{0}c

^{2}strikes a nucleus, an electron – positron pair is created. A positron is an antiparticle of an electron. It is called as a positive electron. As photon is electrically neutral, so in order to converse the charge an electron with a positron must be produced.

E = hf

Or E = hc/λ …………………………………(1) (f = c/f)

Since P = h/λ ……………………………..(2)

So using equs (i) and (ii) we get

E = P.C

Or 0 = F/P

kE

_{max}= hf- φ

Where hf = incident energy

φ = work function

K.E

_{max}= Maximum K.E of photoelectrons

(i). Automatic door system

(ii). Sound strake of movies

(iii). Security system

(iv). Automatic street light

(v). Counting system

^{0}?

Δλ = h/m

_{0}c (1- cosφ)

Where m

_{0}= 9.11 *10

^{31 }kg

h = 6.63 * 10

^{-34}js

θ = 90

^{0}

c = 3*10

^{8}ms

^{-1}

Δλ =

__6.63*10__(1-cos 90

^{-34 }^{0})

9.1 * 10

^{-31}* 3*10

^{8 }Δλ = 2.34 *10

^{-12}m

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