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**MULTIPLE CHOICE QUESTIONS**

**Q.1. A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to(a) H(b) H** (d)

According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.

The wave associated with a moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity. According to de-Broglie theory, the wavelength of de-Broglie wave is given by

Where h = Planck's constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle.

For a body falling freely under gravity from a height H, final velocity of v is obtained by applying kinematic equations

v

or,

We define de-broglie wavelength as

⇒

Where = Constant

⇒ λ ∝ 1/√H. Hence λ ∝ H

(a) 1.2 nm

(b) 1.2 × 10

(c) 1.2 × 10

(d) 1.2 × 10 nm

Ans.

According to Einstein’s quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle being called a photon and possessing energy.

Energy of photon is given by

E = hv = hc/λ; where c = Speed of light, h = Planck's constant = 6.6 x 10

In electron volt,

According to the problem,

Energy of a photon, E = 1 MeV or 10

Now, hc = 1240 eV nm

Now, E = hc/λ

⇒

= 1.24 x 10

(a) No electrons will be emitted as only photons can emit electrons.

(b) Electrons can be emitted but all with an energy, E

(c) Electrons can be emitted with any energy, with a maximum of E

(d) Electrons can be emitted with any energy, with a maximum of E

Ans.

If a beam of electrons of having energy E

The electrons can be emitted with maximum energy E_{0} (due to elastic collision) and With any energy less than E_{0}, when part of incident energy of electron is used in liberating the electrons from the surface of metal.**Q.4. Consider Fig. in the NCERT text book of physics for Class XII. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that**

**(a) Will be larger than the earlier value.(b) Will be the same as the earlier value.(c) Will be less than the earlier value.(d) Will depend on the target.Ans.** (c)

Davison and Germer Experiment:

1. It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of electrons emitted by an . electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron gun.

2. The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it .about the point of incidence. The energy of the incident beam of electrons can also be varied by changing the applied voltage to the electron gun.

According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same but Davisson and Germer found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering. It is maximum for diffracting angle 50° at 54 volt potential difference.

3. If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg’s formula 2d sinθ = nλ, we can determine the wavelength of these waves.

The de-Broglie wavelength associated with electron is where V is the applied voltage.

Using the Bragg’s formula we can determine the wavelength of these waves. If there is a maxima of the, diffracted electrons at an angle θ, then

2d sin θ = A (ii)

From Eq. (i), we note that if V is inversely proportional to the wavelength λ. i.e., V will increase with the decrease, in λ.

From Eq. (ii), we note that wavelength λ is directly proportional to sinθ and hence θ.

So, with the decrease in λ , θ will also decrease.

Thus, when the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that will be less than the earlier value.

4. A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as

The de-Broglie wavelength associated with electron is

...(i)

where V is the applied voltage.

Using the Bragg's formula we can determine the wavelength of these waves. If there is a maxima of the diffracted electrons at an angle θ, then

2d sin θ = λ ...(ii)

From Eq.(i), we note that if V is inversely proportional to the wavelength λ. i.e., V will increase with the decrease in λ.

From Eq. (ii), we note that wavelength λ is directly proportional to sin θ and hence θ.

So, with the decrease in λ, θ will also decrease.

Thus, when the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that will be less than the earlier value.**Q.5. A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as(a) λ _{p} = λ_{n} > λ_{e} > λ_{α} (b) λ_{α} < λ_{p} = λ_{n} > λ_{e} (c) λ_{e} < λ_{p} = λ_{n} > λ_{α} (d) λ_{e} = λ_{p} = λ_{n} = λ_{α}Ans. **(b)

de - Broglie wavelength λ

E

Or λ

M

∴ λ

(a) Remains constant.

(b) Increases with time.

(c) Decreases with time.

(d) Increases and decreases periodically.

Ans.

If a particle is carrying a positive charge q and moving with a velocity v enters a magnetic field 5 then it experiences a force F which is given by the expression

F = q(v x B)=$ F = qvB sin θ. As this force is perpendicular to v and B , so the magnitude of v will not change, i.e. momentum (p = mv) will remain constant in magnitude. Hence,

According to the problem,

Magnetic force on moving electron = -e[v

As this force is perpendicular to so the magnitude of v will not change, i.e., momentum (p = mv) will remain constant in magnitude. Hence, de-Broglie wavelength λ = h/mv remains constant.**Q.7. An electron (mass m ) with an initial velocity v = v _{0}î (v_{0}> 0) is in an electric field E = – E_{0}î(E_{0 }= constant > 0). It’s de Broglie wavelength at time t is given by** (a)

(a)

(b)

(c) λ_{0}

(d) λ_{0}t

Ans.

The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity. According to de-Broglie theory, the wavelength of de- Broglie wave is given by

As initial velocity of the electron is v

λ

Electrostatic force on electron in electric field is,

Acceleration of electron,

Velocity of the electron after time t,

⇒

de-Broglie wavelength associated with electron at time t is λ = h/mv

⇒

⇒

According to the problem de-Broglie wavelength of electron at time t=0.

is λ

Electrostatic force on electron in electric field is

The acceleration of electron,

It is acting along negative y-axis.

The initial velocity of electron along x-axis, v

This component of velocity will remain constant as there is no force on electron in this direction.

Now considering y-direction. Initial velocity of electron along y-axis, v

Velocity of electron after time t along y-axis,

Magnitude of velocity of electron after time t is

⇒

de-Broglie wavelength, λ' = h/mv

⇒

(a) λ = 10 nm

(b) λ = 10

(c) λ = 10

(d) λ = 10

Ans.

de-Broglie wavelength λ = h/mv

For option

(a) λ = 10nm =10 × 10

∴ v = 7.3 × 10

(b) λ=10

= 7.3 x 10

(c) λ = 10

7.3 x 10

(d) λ = 10

v= 7.3 × 10

So the velocity of electron is more for option (c) and (d) where the relativistic correction become necessary although the speed of electron is 7.3 x 10

(a) Their momenta are the same.

(b) Their energies are the same.

(c) Energy of A

(d) Energy of A

Ans.

We know that de-Broglie wavelength λ = h/mv

where, mv = p(momentum) of the particle

⇒ But we can express wavelength

Here, h is Planck constant.

Hence,

But particles have the same de-Broglie wavelength. (λ

Then,

Thus, their momenta is same.

Also,

As p is constant, E ∝ 1/m

∴

Some important characteristics of Matter Waves:

1. Matter wave represents the probability of finding a particle in space.

2. Matter waves are not electromagnetic in nature.

3. De - Brogile or matter wave is independent of the charge on the material particle. It means, matter wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).

4. Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles.

5. Electron microscope works on the basis of de-Broglie waves.

6. The phase velocity of the matter waves can be greater than the speed of the light.

7. Matter waves can propagate in vacuum, hence they are not mechanical waves.

8. The number of de-Broglie waves associated with n

9. Only those circular orbits around the nucleus are stable whose circumference is integral multiple of de-Broglie wavelength associated with the orbital electron.

(a)

(b)

(c)

(d)

de-Broglie wavelength λ = h/p

= 100h/m

Now K.E. = 1/2 mv

Now [from (II)]

⇒ ...(III)

For proton λ

Now [from (III)]

[verifies Ans. (b)]

Now

[verifies Ans. (c)]

(a) Decreases with increasing n, with ν fixed.

(b) Decreases with n fixed, ν increasing

(c) Remains constant with n and ν changing such that nν = constant.

(d) Increases when the product nν increases.

Ans.

We know that energy spent to convert ice into water

E

∴ E = mL = (1000 g) x (80 cal/g)

E = 80000 cal

Energy of photons used = nT x E = nT x hv [∵ E = hv]

So, nThv = mL ⇒ T = mL/nhv

∴ T ∝ 1/n. when v is constant.

Thus, time taken for conversion decreases with increasing v, with n kept constant.

⇒ T ∝ 1/nv

Thus, time taken for conversion decreases with increase in product of nv and T is constant, If nv is constant.

(a) The particle could be moving in a circular orbit with origin as centre

(b) The particle could be moving in an elliptic orbit with origin as its focus.

(c) When the de Broglie wave length is λ

(d) When the de Broglie wavelength is λ

Ans.

According to the question, here given that the de-Broglie wavelength of the particle can be varying cyclically between two values λ

As shown in the figure given alongside,

Let v_{1} and v_{2} be the speed of particle at A and B respectively and origin is at for us O. If λ_{1} and λ_{2} are the de-Broglie wavelengths associated with particle while moving at A and B respectively, then

λ_{1} = h/mv_{1}

and λ_{2} = h/mv_{2}

∴

Since λ_{1} > λ_{2}

∴ v_{2} > v_{1}

By law of conservation of angular momentum, the particle moves faster when it is closer to focus.

From figure, we note that origin O is closed to P than A.

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.14. A proton and an α-particle are accelerated, using the same potential difference. How are the deBroglie wavelengths λ _{p} and λ_{a} related to each other?**

In this problem since both proton and α-particle are accelerated through same potential difference,

We know that,

∴

∴ λ

i.e., wavelength of proton is √8 times wavelength of α-particle.

De-Broglie wavelength associated with the charged particles : The energy of a charged particle accelerated through potential difference V is E = 1/2 mv

Hence de-Broglie wavelength

∴ E

Hv

∴ v

Maximum energy of emitted electron is

E

(ii) The probability of absorbing 2 photons by electron is very low due to their mass difference. So possibilities of such emission of electrons is negligible.

Ans.

But in second case, the energy of the incident photon is low for the substances which has to absorb photons of larger wavelength and energy of emitted photon is high to emit light of shorter wavelength. This means in this statement material has to supply the energy for the emission of photons.

But this is not possible for a stable substances.

Ans.

The photo-electric effect is the emission of electrons (called photo-electrons when light strikes a surface. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of positive ions in the material of the surface.

The photoelectric effect is based on the principle of conservation of energy.

1. Two conducting electrodes, the anode (Q) and cathode (P) are enclosed in an evacuated glass tube as shown on next page.

2. The battery or other source of potential difference creates an electric field in the direction from anode to cathode.

3. Light of certain wavelength or frequency falling on the surface of cathode causes a current in the external circuit called photoelectric current.

4. As potential difference increases, photoelectric current also increases till saturation is reached.

5. When polarity of battery is reversed (i.e., plate Q is at negative potential w.r.r. plate P) electrons start moving back towards the cathode.

6. At a particular negative potential of plate Q no electron will reach the plate Q and the current will become zero. This negative potential is called stopping potential denoted by V_{0}. Maximum kinetic energy of photo electrons in terms of stopping potential will therefore be K_{max} = (|V_{0}|) eV

So we conclude that in photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.

Therefore, all the electrons that absorb a photon doesn't come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of metal.**Q.18. There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength?Ans.**

1. X-rays were discovered by scientist Roentgen that is why they are also called Roentgen rays.

2. Roentgen discovered that when pressure inside a discharge tube is kept 10

3. There are three essential requirements for the production of X-rays.

(i) A source of electron

(ii) An arrangement to accelerate the electrons

(iii) A target of suitable material of high atomic weight and high melting point on which these high speed electrons strike.

Here in this problem total energy will be constant.

Let us assume wavelength of X-rays is λ

Given, P = 100 W

λ

and λ

Let n

So,

⇒

⇒

**SHORT ANSWER TYPE QUESTIONS**

**Q.19. Consider Fig. for photo-emission.**

**How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.****Ans. **The momentum of incident photon is transferred to the metal ,during photo electric emission.

At microscopic level ,atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.The excited electron is emitted. Therefore, the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electron.**Q.20. Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.Ans. **Let the maximum energies of emitted electrons are K

k

( ∴ hc = 1240 ev nm)

Work function φ = 6.2/6 = 1.03 eV.

Ans.

Therefore,

⇒

= 1.05 x 10

Energy, [∵ p = Δp]

⇒

= 3.8 x 10

Ans.

n

∵ n

∴ 2n

2v

So the frequency of source B is twice the frequency of source A.

Ans.

|p

Let us first take the case I when both p

In second case when both p

In case III when p

⇒

And in case IV when p

∴

⇒

Ans.

2d sinθ = nλ

n = 1 so λ = 2d sinθ [ θ = 30°(Given)]

= 2 × 0.1 × 10

= 6.6 x 10

**LONG ANSWER TYPE QUESTIONS**

**Q.25. Consider a thin target (10 ^{–2}m square, 10^{–3}m thickness) of sodium, which produces a photocurrent of 100µA when a light of intensity 100W/m^{2} (λ = 660nm) falls on it. Find the probability that a photoelectron is produced when a photons strikes a sodium atom. [Take density of Na = 0.97 kg/m^{3}].**

Thickness (d) = 10

Current (i) = 100 μA = 10

Intensity (I) = 100 W/m

Mass of target (m) = Vol. × density

m = Area of sheet × thickness × density

= (10

M = 0.97 ×10

∴ No. of Na Atoms in target =

Number of Na atoms in Na target = 2.54 × 10

Total energy falling per second on target = nhv

Intensity × Area = n x h x = c/λ

Number of photons (n) incident per second on Na-metal

∴ n = 3.3 × 10

Let P is the probability of emission of photo-electrons per atom per photon.

Number of photo-electrons emitted per second

N = P·n. (No. of sodium atom)

N = P × 3.3 × 10

i = 100μA = 10

⇒ N = i/e

P = 7.5 × 10^{-21}**Q.26. Consider an electron in front of metallic surface at a distance d (treated as an infinite plane surface). Assume the force of attraction by the plate is given as Calculate work in taking the charge to an infinite distance from the plate. Taking d = 0.1nm, find the work done in electron volts. [Such a force law is not valid for d < 0.1 nm].Ans.** Consider the figure in which an electron is displaced slowly by a distance x by the means of an external force which is,

where, d = 0.1 nm = 10

Let the electron be at distance x from metallic surface. Then, force of attraction on it is

Work done by external agency in taking the electron from distance d to infinity is**Q.27. A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V _{stop} vs ν is given in Fig.**

**(i) Which material A or B has a higher work function?(ii) Given the electric charge of an electron = 1.6 × 10 ^{–19} C, find the value of h obtained from the experiment for both A and B.Comment on whether it is consistent with Einstein’s theory:Ans.** Threshold frequency (n

If incident frequency n < n

For most metals the threshold frequency is in the ultraviolet (corresponding to wavelengths between 200 and 300 nm), but for potassium and cesium oxides it is in the visible spectrum (λ between 400 and 700 nm).

Here we are given threshold frequency of A

v

For B, v

We know that

Work function,

⇒ Ф

So,

⇒ Ф

Thus, work function of B is higher than A.

(ii) For metal A, slope = h/e =

or

= 6.4 x 10

For metal B, slope = h/e =

Or

= 8 x 10

Since, the vale of h from experiment for metals A and B is different. Hence, experiment is not consistent with theory.

Ans.

By the law of conservation of momentum,

m

⇒ m

And the law of conservation of energy,

...(i)

⇒

⇒ ...(ii)

Dividing Eq. (ii) by Eq. (i),

we get v + v

Solving Eqs. (i) and (iii), we get

(i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses]

(ii) Will there be photoelectric emission?

(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?

(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?

(v) Can you explain how photoelectric effect was observed instantaneously?

[Hint: Time calculated in part (iii) is from classical consideration and you may further take the target of surface area say 1cm2 and estimate what would happen?]

Ans.

r = 1.5 Å = 1.5 × 10

Let number of photons emitted by bulb per second is n

P = n

n

Number of photons emitted by bulb per second

n

(ii) Energy of incident photon E = hv = hc/λ

= 2.48 eV

Energy of photon = hv = hc/λ =

Energy of photon

As the energy of an incident photon is more than 2 eV i.e., the work function of metal surface hence the photoelectric emission takes place.

(iii) Let ΔT time spent in getting the energy φ (work function of metal) Energy received by atomic disk in

Δt time E = P × A· Δt

E = P × πr^{2}·Δt

Energy transferred by bulb in full solid angle 4πd^{2} to atoms = 4πd^{2}φ

(iv) Number of photons received by one atomic dick in time Δt is

[n_{1} from part (i) and Δt from part (iii)]

photon per atom

N = 1 photon per atom.

(v) Time of emission of electrons is 11.4 sec. So the photoelectric emission is not instantaneous in the problem. It takes about 11.4 sec.

In photoelectric emission there is a collision between incident photon and free electron and nucleus, which lasts last for very - very short interval of time (10^{-9 }sec) hence we say photoelectric emission is instantaneous.

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