IIT JAM Physics MCQ Test 17


60 Questions MCQ Test Mock Test Series for IIT JAM Physics | IIT JAM Physics MCQ Test 17


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This mock test of IIT JAM Physics MCQ Test 17 for Physics helps you for every Physics entrance exam. This contains 60 Multiple Choice Questions for Physics IIT JAM Physics MCQ Test 17 (mcq) to study with solutions a complete question bank. The solved questions answers in this IIT JAM Physics MCQ Test 17 quiz give you a good mix of easy questions and tough questions. Physics students definitely take this IIT JAM Physics MCQ Test 17 exercise for a better result in the exam. You can find other IIT JAM Physics MCQ Test 17 extra questions, long questions & short questions for Physics on EduRev as well by searching above.
QUESTION: 1

How much energy must be given to an election to accelerate it to 0.95c?

Solution:

Use the mass-increase formula.


E = (m - m0)c2
= [(29.1 - 9.1) (10-31)] (9 x 1016
= 1.8 x 10-13 J = 1.125 MeV

QUESTION: 2

The de-Broglie wavelength of a proton and an α - paiticle are equal. The ratio of their velocities is:

Solution:

Every particle of mass m moving with velocity v is associated with a wave of wavelength given as
 Planck's constant
The wavelength for proton

and the wavelength for He

Since,

⇒  
Since 2He4 has four nucleons so m = 4mP
Thus, (iii) 

QUESTION: 3

Let the potential energy of a hydrogen atom in the ground state be zero. Then its energy in the first excited state will be

Solution:

Since we assume the potential energy to be zero in the ground state . So Total energy = U + K = Kinetic energy 
Since P E = U = 0
∴ Total energy = 13.6ev (in ground state)
If the potential energy is not assigned a zero value, then the total energy is - 13.6ev .
So we conclude that making potential energy zero increases the value of total energy by 13.6 - ( - 13.6) = 27.2ev 
Now actual energy in second orbit = - 3.4ev
Hence new value is ( - 3.4 ev + 27.2 ev) = 23.8ev

QUESTION: 4

A clock keeps correct time, with what speed should it be moved to an observer so that it may appear to loose 5 minutes in 24 hours?

Solution:

We use 
where is proper time = 1440 minutes in a day and t = 1445 minutes measured by a stationary obseiver.
So 
So 
or 
 

QUESTION: 5

The rate of radiation of a black body at 00C is E Watt. Then the rate of radiation of this black body at 2730C will be :

Solution:

By Stefan's law, rate of emission is proportional to fourth power of the absolute temperature i.e.,

⇒  ⇒ 
Here.  T1= 0 + 273 = 273. T2 = 273 + 273 = 2 x 273
SO, 

QUESTION: 6

A harmonic oscillator has the wave function   is the eigenfunction belonging to the nth energy eigenvalue  The expectation value <E> of energy for the state ψ(x,t) is

Solution:

 for Harmonic oscillator energy eigen value is 





QUESTION: 7

Consider a quantum mechanical system with three linear operators  which are related by  Where  is the unit operator. If  the  must be 

Solution:



QUESTION: 8

Electric field component of an electro magnetic radiation varies with time as  where a is a constant and the values of the ω and  respectively. This radiation falls on a metal of work function 2 eV. The maximum kinetic Energy of photoelectrons is.

Solution:



The maximum frequency of photon = 

QUESTION: 9

The electrons x and y are moving with velocities 0.1 C and 0.8C respectively. The ratio of energies ΔEX / ΔEy required to increase the speed of each by 0.1 C is given approximately by

Solution:





QUESTION: 10

A particle A of mass mA decays into two particles, one of which is B with mass mB and the other massless. The magnitude of the momentum of particle B using relativistic kinematics in the first-frame of the particle A is

Solution:


According to law of conservation of linear momentum. The momentum of particle B and massless particle are equal in magnitude and opposite is sign. Let their momentum be p.


Momentum of particle B is rest frame of particle a is same as that of massless particle

QUESTION: 11

A driver is being fined for overlooking a red signal. The driver claimed that instead of red colour (v0 = 4.8 * 1014 Hz) he saw green (v = 5.6 * 1014 Hz) because of Dopper effect. The judge fined the driver for speed at the rate of Rs. 1/- for each km/h he exceeded the speed limit of 80 km/h. What is the fine?

Solution:


 = 4.59 107 m /sec. = 1.65 x108 km/h
1 m/sec. = 3.6 km/hr
fine = 1.65 x 108 - 80 = 164999920/-

QUESTION: 12

In a Devisson - Germer experiment a collimated beam of electrons of energy 54eV at a normal incidence on a given crystal, shows a peak appears at the same value of reflection angle, then the energy of the neutrons must be

Solution:

In Davission - Germer experiment when electron is replaced by neutron, then same peak at a reflection angle of 40° is obtained which shows the same wavelength of e- and neutron.



QUESTION: 13

The photo electric threshold of certain metal is 2750 A°. Find the maximum kinetic energy of the electrons. 

Solution:

Work function 

Maximum kinetic energy

QUESTION: 14

The velocity of an electron in the ground state of a hydrogen atom is VH. If Vp be the velocity of an electron in the ground state of positronium, then

Solution:

H-atom having one electron in its ground state. So. having VH velocity and positronium is the electron category, that is a particle having mass equal to electron but charge opposite in nature and same magnitude. 
qp = 1.6 x 10-19
Mp = 9.1 x 10-31kg
So, velocity of positronium (Vp) is equal to the velocity of electron in H-atom VH
Vp = VH

QUESTION: 15

A beam of atoms moving in a certain direction can be slowed down if they absorb photons from a laser beam moving in the opposite direction and subsequently spontaneously emit photons isotropically. For a beam of sodium atoms (mass number A = 23) with speed 600 m/sec. If a laser beam of wavelength 589 nm is used the number of such absorption and emission cycles needed to bring a sodium atom to rest would be approximately________

Solution:

Photon momentum P = h/λ (for one cycle)
P must be equal to mV.
If n is no. of cycles then

m = 23 x 1.66 x 10-27 kg = 3.818 x 10-26kg 
V = 600 m/sec 


n = 2.1 x 104

QUESTION: 16

An α-particle and a proton are fired through the same magnetic field which is perpendicular to their velocity vectors. The α-particle and the proton move such that radius of curvature of their path is same. What is the ratio of their de-Broglie wavelengths?

Solution:

Magnetic force experienced by a charged particle in magnetic field is given by

 ⇒ mv = qBr

 

QUESTION: 17

In a collision between a photon and a free electron, which is correct, if all the photon energy were transferred to the electron.

Solution:

Let E1 = hv energy of photon
 momentum of photon
 energy of electron
P2 = momentum of electron 
If all energy is to be transferred to the electron hv 
Electron momentum

QUESTION: 18

Denote the computer of two matrices A and by [A, B] = AB - BA and anti commutator by {A,B} = AB + BA 
If {A.B} = 0 we can write [A,B,C]

Solution:

Given that {A, B} = 0 or AB = - BA 
⇒ [A.B.C] = [A.B] C + B [A.C] = ABC - BAC + BAC - BCA 
= ABC - BCA = - BAC - BCA = - B (AC + CA) 
[A.BC] = - B {A.C}

QUESTION: 19

The activity of a radioactive sample is decreased by 25% of the initial value after 30 days. The half - life (in days) of the sample is approximately

Solution:




Half life 
Half life  

QUESTION: 20

Five identical non-interacting particles, each of spin 1/2 and mass m, are moving in a one-dimensional infinite potential well given by
 
The energy of the lowest energy state is

Solution:


Energy of the lowest energy state of system


QUESTION: 21

The frequency of electron in Bohr orbit is given by f1 revolutions/sec. The frequency of electron in the n-th orbit for n > 1 is

Solution:

We know that
Radius of Bohr atom is 

and velocity of electron

Time period of cycle


T α n3
⇒ frequency of electron = f1/h3

QUESTION: 22

An electron is moving inside a uniformly charged sphere with total charge +ve. The uncertainty principle ΔPxΔx ≥ n is used for finding the smallest radius rmin of the charged sphere such that the kinetic energy of the electron is equal to its potential energy on the surface of the sphere. The rmin is approximately in the range    .

Solution:

Potential energy of interaction between electron and sphere






QUESTION: 23

The Compton effect experiment photons of energy h v around material of atomic number Z. The change in wavelength can be

Solution:

The change in wavelength

= 0.0243(1 - cos φ) A0
Therefore the change is easily detectable in case of X-ray (wavelength ≈ 1 A0)

QUESTION: 24

1.5 mW of 400 nm light is directed at a photo electric cell. If 0.1 % of the incident photon produce e-s. all the photo e- s reach the opposite plate, so the current in the photo cell is equal to______ .

Solution:

Since  ...(i)
⇒ n/t ⇒ no. of photo e-s produced per second.
Now let the no. of photons hitting the photo cell per second will be n 
nhv = 1.5 x 10-3 Watt


No. of photo e- produced per second is

= 3 x 1011 e/sec.
Using equation (1)
I = 3 x 1011 x 1.6 x 10-19
I = 4.8 x 10-7 amp = .48 μ amp.

QUESTION: 25

In a hydrogen - like atom, an election is the ground state requires 476 ev to reach an excited level with quantum number 2n. If it makes a transition from this level to a lower level with quantum number n, it emits a photon of energy 40.8 eV. Find the atomic number Z of the element

Solution:

For hydrogen like atom 

Now. it requires 476 eV to reach 2n level from ground state 
E2n- E1 = 476
 ...(i)

It emits energy of 40.8 eV while making transition from 2n level to n level 
E2n - En = 40.8

 ...(ii)
Dividing (1) and (2)

n = 3
Putting n = 3 in (2) we get z = 6

QUESTION: 26

The speed of an electron, whose de Broglie wavelength is equal to its Compton wavelength, is (C is the speed of light)

Solution:



QUESTION: 27

A neutron of mass mn = 10-27 kg is moving inside a nucleus. Assume the nucleus to be a cubical box of size 10-14 m with impenetrable walls. Take Js and 1 MeV ≈ 10-13 J . An estimate of the energy in Mev of the neutron is

Solution:





QUESTION: 28

The longest wavelength in the Balmer series of hydrogen atom is given as 656 nm positronium can be considered as a system of an electron and a positron in which both these particles orbit each other. What will be the longest wavelength in the lyman series for positronium?

Solution:

The expression of wavelength emitted when an electron makes a transition from nth level to mth level is given by

for Balmer series m = 2

The value for n for longest wavelength will be 3



for positronium mass m will be replaced with reduced mass μ. 

Rydberg constant for H - atom

Rydberg constant for positronium

for lyman series 
for longest wavelength n = 2


QUESTION: 29

Positronium is a hydrogen like bound state of a position and electron. If the nth energy level of hydrogen atoms is given by  the nth energy  of the positronium will be equal to

Solution:

Mass of e+ (positron) = mass of e- (electron) = m
so, Reduced mass of positronium 
m = m/2
so, if 


QUESTION: 30

A particle has rest mass m0 and momentum m0c, where c is the velocity of light. The total energy and the velocity of the particle are respectively.

Solution:

That energy of a relativistic particle is given by
E2 = p2C2 + m20c

Again, momentum of the particle 
⇒ 
⇒ 

*Multiple options can be correct
QUESTION: 31

The initial state of a system is given in terms of a complete and orthonormal basis that has four vectors

and the eigenvectors to the Hamiltonian with energies E1 E2, E3 and E4 respec­tively.

Solution:

For the normalizing function


Then the probabilities of the system 

 
at t = 0

*Multiple options can be correct
QUESTION: 32

A particle is moving in a two - dimensional potential well

Which of the following statements about the ground state energy E1 and ground state eigenfunction ψ0 are true?

Solution:

We know that



Then 

*Multiple options can be correct
QUESTION: 33

For the three Pauli matrices, which one of the choices are incorrect?

Solution:

We know that 

So, By taking the option

*Multiple options can be correct
QUESTION: 34

Cosmic ray muons generated at a top of the earth atmosphere decay law.

{ Given : T1/2 of muons = 1.92 μ sec , velocity v = 0.96 c }
An observer from the top of the mountain of height 2 km about mean sea level detect muons with their speed of light and count 1000 muons, the no. of muons (n) of the same speed detected by an observer at mean sea level at the same period of the time would be n, then

Solution:


Now the time taken by the rays to reach the sea level

Again we know that

No = 1000
N(t) = 1000 exp 
= 495

*Multiple options can be correct
QUESTION: 35

In an one - dimensional harmonic oscillator , are respectively the ground, second and forth excited states. These three states are normalized and are orthogonal to one another.
ψand ψ2 are two states defined by

Solution:


⇒ 1 + 2 + 3α = 0
α = -1



 

*Multiple options can be correct
QUESTION: 36

X - Rays of 8 wavelength falls on election cloud and gets scattered. Then

Solution:

Maximum change in K.E. will correspond to maximum change in wavelength
compton shift 

KE|max of electron 



We know that the Rest mass energy of electron = 0.51 M eV = 5.1 x 105 eV K.E. of electron <<< Rest mass energy

= 0.0569 x 3.162 x 107
= 1.8 x 106 m/sec 

*Multiple options can be correct
QUESTION: 37

In the hydrogen atom spectrum, which of the following statements are correct?

Solution:




*Multiple options can be correct
QUESTION: 38

If the normalized wavefunction for a particle in a one dimensional box is given by
ψ{x) = A (x2 -1) for -1 ≤ x ≤ 1 then

Solution:

Applying normalizing condition 



*Multiple options can be correct
QUESTION: 39

For the Pauli matrix 

Solution:


Eigen value equation 

Now for the eigen value ±1 , the eigenvectors are

*Multiple options can be correct
QUESTION: 40

A real operator   satisfies the lowest order equation

Solution:

The given equation is 

And eigen state of A are 

*Answer can only contain numeric values
QUESTION: 41

De-Broglie wavelength associated with the hydrogen atom moving with most probable velocity at 270C is 1.26De-Broglie wavelength associated with helium atom moving with r.m.s velocity at 510C is equal to_____  .


Solution:

We know that most probable velocity = 

*Answer can only contain numeric values
QUESTION: 42

A radioactive source contains two radioisotopes, each with initial activity 103 In 2 Bq (1 Bq = 1 decay per second). The half - life of one of the radioisotopes is one hour and that of the other is two hours. Calculate the total number of radioactive nuclei present initially in the radioactive source is __________ x 107.


Solution:

Half life 
let half of two radio isotopes are t1 and t2 & decay constants are λ1 and λ2.
Let initial number of radioactive nuclei present in the sample are N10 and N20 Activity of radioactive material is given by

Let activity of radioactive materials are A1 and A2.


Initial activity 


So, total number of radioactive nuclei initially present in the radioactive source

*Answer can only contain numeric values
QUESTION: 43

A particle is initially in its ground state in an infinite one - dimensional potential box with sides at x = 0 and x = a. If the wall of box at x = a is suddenly moved to x = 3a, calculate the probability of finding the particle in the ground state of the new box is ________(in percentage).


Solution:

When particle is initially in ground state. The state of the system is defined by

When the wall of the particle is moved to x = 3a the state of the system will remain same but eigenstates of the system will change Now, the eigenstates are defined by eigenfunction



probability of finding the particle in ground state is given by







So, probability of finding the system in ground state is given by

*Answer can only contain numeric values
QUESTION: 44

A pion (rest mass m0 = 135 MeV/c2) is moving with a velocity  If it decays by emitting two photons both of which move in z-direction, find their energies in unit of MeV.


Solution:

Let us attach frame S' with pion In S’ frame, pion is at rest and it decays into two photon.
According to conservation of linear momentum, each photon will have equal & opposite momentum of each photon is given by h/λ

So, they will have equal wavelength and hence equal frequency. According to observation of energy hv + hv = m0 c2 = 135 MeV.

for an observer in s frame, the source of radiation moves towards right with velocity 0.8 c
In S frame 


So, in S frame, frequency of each photon will be 
Associated energy E = hv’ = 22.5 MeV

*Answer can only contain numeric values
QUESTION: 45

Cosmic ray muons generated at the top of the Earths atmosphere decay according to the radioactive decay law

Where N(t) is the number of muons at time t and T1/2 = 1.52 μsec is the proper half - life of the muon. Immediately after generation, most of these muons shoot down towards the earths surface. Some of these muons decay on the way but their interaction with atmosphere is negligible. An observer on the top of a mountain of height 2 km above mean sea level detects muons with the speed 0.98 cover a period of time and counts 1000 muons. The number of muons of the same speed detected by an observer at mean sea level in the same period of time would be____.


Solution:



It is given that T1/2 = 1.52 μsec
N(t) = N(0) exp 

N(0) = 1000

N(t) = 1000 exp (-0.617214) 
N(t) = 539

*Answer can only contain numeric values
QUESTION: 46

A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?


Solution:

The intensity of light at a distance of 2m from the source

and as the area of target π x 10-28m2 the energy absorbed by the electron per sec.

So, the time taken by electron to absorb 5 eV energy

*Answer can only contain numeric values
QUESTION: 47

A Photo sensitive material em its electron whose maximum velocity is 3 x 107ms-1. if mass of e- = 9.1 x 10-31 kg and charge e = 1.6 x 10-19c, then what is the value of stopping potential (in V)?


Solution:

We know that 


25.59 x 102 
2559

= 25,59 x 102volt 
= 2559 volt

*Answer can only contain numeric values
QUESTION: 48

Two spherical nuclei have mass numbers 64 and 8 with their radii R1 and R2 respectively. Then calculate the ratio of R1 to R2.


Solution:

We know radius of nuclei is 

given for one nuclei 
A1 =64
and other A= 8
So, 

*Answer can only contain numeric values
QUESTION: 49

A beam of X-Ray of wavelength 0.2 nm is incident on a free electron and gets scattered in a direction with respect to the direction of the incident radiation resulting in maximum wavelength shift. Then find out percentage energy loss of the incident radiation.


Solution:

Compton shift 
For maximum shift 

Energy of radiation

log E = log hc - logλ
taking derivative

Percentage energy loss 

*Answer can only contain numeric values
QUESTION: 50

The nuclear reaction observed to occur even when very slow moving neutrons (Mn = 1.00874) strike a boron atom at rest. For a particular reaction in which K≈ 0, the helium (MHe = 4.00264) is observed to have a speed of 9.3 x 106 m/sec. Determine the Q - value (MeV) of the reaction { MLi = 7.0160 u)


Solution:

Since the neutron and boron are both initially at rest the total momentum before the reaction is zero and afterward is also zero.
Therefore MLi VLi = MHe VHe
We solve this for VLi and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistic formulas, be-
2.82
Since the neutron and boron are both initially at rest, the total momentum before the reaction is zero and afterward is also zero.
Therefore MLi VLi = MHe VHe
We solve this for VLi and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistic formulas, because VHe = 9.30 x 106 m/sec is not close to the speed of light c and VLi will be even less Since MLi > MHe. Thus we can write

We put in numbers, changing the mass in u to kg and recalling that 1.6 x 10-13 J = 1 MeV
 
Given data Ka = Kx = 0 
So Q = KLi + KHe

*Answer can only contain numeric values
QUESTION: 51

A K meson (with a rest mass of 494 MeV) at rest decays into a muon (with a rest mass of 106 MeV) and a neutrino. Then energy of the neutrino, which can be taken to be massless. (In MeV).


Solution:






Ev = 236 MeV

*Answer can only contain numeric values
QUESTION: 52

If a beam of electrons impinges on an energy barrier of height 0.030eV and of infinite width, find the fraction of electrons reflected at the barrier if the energy of the impinging electrons is 0.040 eV.


Solution:

Given E = 0.040 eV = 0.040 x 1 .6 x1 0-19J = 0.064 ,10-19J
V0 = 0.030 eV = 0.030 x 1.6 x 10-19 J
E>V0
Then the fraction of electrons reflected, the reflectance is given by

We know

P1 = 1.07x 10-25 kg. m/sec


Reflectance 
Reflectance = 0.1

*Answer can only contain numeric values
QUESTION: 53

The de Broglie wavelength of a proton of energy Ep is twice the de Broglie wavelength of an alpha particle of energy Eα, then find out the ratio of 


Solution:


*Answer can only contain numeric values
QUESTION: 54

The proper lifetime o f an unstable particle is equal to Δt0 = 10 ns. Find the distance this particle will traverse till its decay in the laboratory frame of reference, where its lifetime is equal to Δt0 = 20 ns (in m).


Solution:

The distance traveled in the laboratory frame of reference is vΔ where v is the velocity of the particle, But by time dilation

Thus the distance traversed is

S = 5 m

*Answer can only contain numeric values
QUESTION: 55

Two particles moving in a laboratory frame of reference along the same straight line with the same velocity v = 3/4 C strike against a stationary target with the time inteival Δt = 50 ns. Find the proper distance between the particles prior to their hitting the target (in m).


Solution:

In the frame K in which the particles are at rest, their positions are A and B whose coordinates may be lopen as A : (0,0,0), B = (lo, 0,0)
In the frame K’ with respect to which K is moving with a velocity v, the coordinates of A and B at the time t’ in the moving frame are


Suppose B hits a stationary target in K' after time while A hits it after time tB+Δ t Then


l0= 17 m

*Answer can only contain numeric values
QUESTION: 56

The expectation value of p* 2 for the wave function  in the region and  Then the value of n is ______.


Solution:

The expectation value of observable p in state ψ is given by 

where is the operator as sociated with x component of momentum is
The expectation value of p2 in state ψ is given by


*Answer can only contain numeric values
QUESTION: 57

Consider a one - dimensional harmonic oscillator of angular frequency ω. If 5 identical electrons occupy the energy levels of this oscillator at zero temperature,
Then the ground state energy E0 = ______


Solution:

For one-dimensional harmonic oscillator 

*Answer can only contain numeric values
QUESTION: 58

A particle of mass m and energy E, moving in the positive x-direction, encounters a one dimensional potential barrier at x = 0. The barrier is defined by 
V = 0 for x < 0
V = V0 for x > 0 (E > V0)
If the wave function of the particle in the region x < 0 is given as Aeikx + Be-ikx
If B/A = 0.4, then find E/V.


Solution:









*Answer can only contain numeric values
QUESTION: 59

In a compton experiment, the ultraviolet light of wavelength 2430 A0 is scattered from an electron at rest. What should be minimum resolving power of an optical instrument to measure the compton shift, If the obseivation is made at 1800 with respect to the direction of the incident light.


Solution:

for uv light of wavelength 2430 A0 scattered at 1800

for θ = 1800

= 0.0484 A0
minimum resolving power of instrument required to measure the compton shift

R = 50206

*Answer can only contain numeric values
QUESTION: 60

An “excited" atom gives up its excess energy by emitting a photon of characteristic frequency. The average period that elapse between the excitation of an atom and the time it radiates is 0.75 x 10-8 sec. Calculate the inherent uncertainty in the frequency of the photon? (in MHz)


Solution:

The photon energy is uncertain by the amount

and the corresponding uncertainty in the frequency of light is

= 10.6 x 106 Hz = 10.6 MHz = 11 MHz.

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