Free Objective Test 01 Practice Test - 11th and 12th
Question 1
Why is exposure to X- ray injurious to health but exposure to visible light is not, when both are electromagnetic waves?
SOLUTION
Solution : B
X-rays are a part of electromagnetic spectrum and they fall in the high energy region of it.
They are not radioactive neither charged, but it is only them being highly energetic that they may affect the chemical bonds inside the body and change certain mechanisms if not destroy them (simply speaking creating mutation in fundamental building blocks of human body as D.N.A.). Medical science calls these types of mutations "Cancer".
That’s why they are harmful.
But don't get scared of becoming some mutilated character next time when you go for an x-ray examination!! The chances of this are very less and the benefits are definitely more (your doctor gets to see your fractures). The cancer may be caused only due to excess exposure to direct radiation.
Question 2
The x-ray target was put at an angle to the incident electrons so that produced x-rays can be collected in particular direction. How about instead we put a magnetic field around the target that will bend the x-ray in our preferred direction?
SOLUTION
Solution : B and D
Magnetic field can only influence a magnet and hence moving charges (Because they are magnets too). x-ray on the other hand is not a charged particle but a packet of electro-magnetic energy. Hence bending x-rays using a magnetic field is out of the equation, moreover even if tried so, a presence of magnetic field could affect the path of electrons present in the tube. Won't it ?
Question 3
The Kα x-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 23.32 keV, what will be the energy of this atom when an L electron is knocked out? If required, take hc = 1242 eV.nm.
SOLUTION
Solution : B
In our notation of energy levels, an energy EX will mean the energy of the atom when there's a missing electron in the X shell ( X = K, L, M, N,....). Writing quantitatively.
E(Kα) = EK−EL
⇒EL=EK−E(Kα)
⇒EL=EK−(hcλKα)
⇒EL=23.32keV−(1242eV.nm0.071nm)=23.32keV−17.5eV
⇒EL=5.82keV.
Question 4
X−Kα
Y−Kβ
Z−Kγ
SOLUTION
Solution : A
This is a definition based question.
Kα : When electronis transition takes place from L-shell to K-shell or vacancy shifts from K-shell to L-shell (this is what the energy level diagram indicates)
We defined as following :
Kβ : When electronis transition takes place from M-shell to K-shell or vacancy shifts from K-shell to M-shell (this is what the energy level diagram indicates)
Kγ : When electronis transition takes place from N-shell to K-shell or vacancy shifts from K-shell to N-shell (this is what the energy level diagram indicates)
Got the answer!!
Question 5
There is a vacancy in the L shell. Which of these transitions are possible?
SOLUTION
Solution : B and C
So we have an electronic vacancy created in L-shell. Now let's think about each cases one by one. We just have to keep in mind that every system acts in a way to minimize its energy. No process is spontaneously possible in which the energy of the system increases. Only those transition/s will be possible after which the energy of the atom decreases.
From the atomic energy level diagram we know that the atomic energy is highest when there is a vacancy in K-shell and decreases further. i.e.
EK > EL > EM > ENAnd so on.
If a transition from K-shell occurs, the vacancy will shift to K-shell. The atomic energy increase hence. So, this is not possible.
If a transition from M-shell happens, the atomic energy decreases, hence such a transition is possible.
Infact any transition in which the electron is jumping from a shell above L-shell to fill a vacancy here is possible.
Question 6
Arrange the given atoms in ascending order of their energy.
1)
II)
III)
IV.
SOLUTION
Solution : D
It’s a very simple question. We have to compare the energy in all the four diagrams. Basically these diagrams represent an atom with different electronic configuration with, infact, an electron missing from a specific shell in each case. Proceeding organically, first of all let's say the energy in the first case (with each electron at its place) is x.
Now, in the second diagram the electron is removed from the K-shell. We need to supply energy to the atom to do this right? So let's say we supplied an energy ΔK.
So, the atom's energy now becomes x+ΔK.
Now in the third diagram the electron is removed from M shell. We need to supply energy even for that. Let's say that energy is ΔM.
So, atom’s energy becomes x+ΔM.
In the fourth diagram, similarly after removing electron from L-shell the atomic energy becomes x+ΔL.
Now, we understand that it is tougher to remove the electron which is closer to the nucleus as it is more strongly bounded with the nucleus. To remove the electron from K-shell we need to supply the highest amount of energy, and to remove an electron from the M-shell we need to supply the least amount of energy. Mathematically,ΔK > ΔL > ΔM
Hence the energy of the atom will be in the order: II > IV > III > I.
Question 7
The wavelength λ of Kβ X-Ray is given by:
SOLUTION
Solution : A
Kβ characteristic X-Ray phton is emitted when an electron from the M shell jumps to fill the electronic vacancy created in K shell.
Now, if we refer to the Energy Level Diagram, we find that The atomic energy when there is a vacancy in K shell is EK.
Also the atomic energy when the vacancy gets created in M shell is EM.
We see that EM < EK.
So the atom has lost this energy. How much ?
EK−EM
This energy appears as the energy of Kβ characteristic X-Ray phton.
Hence hcλKβ=EK−EM.
Which gives,
1λKβ=EK−EM.
Question 8
The Kα x-ray emission line of tungsten occurs at λ = 21 pm. What is the energy difference between EK and EL levels in this atom (MeV refers to mega electron volts;1MeV = 106 eV and hc = 1242 eVnm)?
SOLUTION
Solution : A
Kα x-ray emission takes place when an electron jumps from L-shell to fill a vacancy created in K-shell.
Now EK is the energy of the atom when vacancy is in K-Shell, and EL Is the energy of the atom when vacancy is created in L-Shell. Also, EK > EL So, when vacancy shifts from K to L, according to our agrument, the atomic energy decreases. How much ?
EK−EL.
This difference results into X-ray photon.
Hence connecting the dots we can conclude that ; The energy difference between K and L levels is the energy of the photon.
Mathematically;
EK−EL=Ephoton
Now,
EK−EL=hcλ
Which gives after putting in the values;
EK−EL=1242 eVnm0.0021nm
Final calculations leads us to our answer;
EK−EL=591428eV
Or,
EK−EL=0.59MeV
Question 9
The wavelength of Kα and Lα X-Rays of a material are 20 pm and 140 pm respectively. Find the wavelength of Kβ X-Ray of the material.
SOLUTION
Solution : C
For the wavelength of Kα x-rays; hcλ1 = EK−EL
for the wavelength of Lα x-rays; hcλ2 = EL−EM
or the wavelength of Kβ x-rays; hcλ3 = EK−EM
Now,
If we add the first two equation we get :
hc(1λ1+1λ2)=EK−EM
Now from equations iii) and iv)
hcλ3=hc(1λ1+1λ2).
We get;
1λ3=(1λ1+1λ2).
Or,
1λ3=(120+1140)pm−1
Or,
λ3=2800160pm.
Or,
λ3 = 17.5 pm
Question 10
Any energy level marked here in the diagram can best be:
SOLUTION
Solution : B
The energies marked in the diagram are the energies of the atom when a vacancy of electron is created in particular shells.
For example, the energy marked as EK is the energy of atom when one electron from the K shell is removed. And so on.
It is interesting to note that the energy of the atom with one vacancy in K shell is higher; and as we move towards the higher shells, the energies of the atom with vacancies in those shells decrease. Also, the energy is zero when the atom is in its ground state.
This can be understood simply, as the K shell electrons are bound to the nucleus most strongly; this force decreases as we move to higher shells. Hence, to remove an electron from the lowermost shell (K) maximum amount of energy is required to be supplied, this amount will decrease as we move towards higher shells; for the same reason.
In the ground state (no electrons removed) the atom’s energy will be the least, which is taken to be zero.
Question 11
The intensity of x-rays from Coolidge tube is plotted against wavelength λ for a molybdenum target at an accelerating voltage of 35 kV. The minimum wavelength found is λ0 and the wavelength of the Kα line is λKα. As the accelerating voltage is increased from 35 kV to 70 kV -
SOLUTION
Solution : A
The Kα peak corresponds to radiation when the atom transitions between two energy levels due to an internal jump of an electron from the L shell to the K shell. It is characteristic for a particular element and will not depend on any external parameters, like the accelerating potential.
But increasing the accelerating potential from 35kV to 70kV will now allow x-rays carrying energies in the range 35-70 eV, which was not allowed earlier. There will now be a higher cut-off on the energy, and hence, a lower cutoff on the wavelength (since E∞1λ). The new cut-off wavelength λ0 will be lower than λ0, and farther from λKα.
Thus, among the given options, the only thing that will change when you increase the accelerationg potential is that (λKα−λ0) will increase.
Question 12
For a given material, the energy and wavelength of characteristic x-rays satisfy:
SOLUTION
Solution : C and D
Quantitatively,
E(Kα)=EK−EL
E(Kβ)=EK−EM
E(Kγ)=EK−EN
Clearly,
E(Kα) < E(Kβ) < E(Kγ). (1)
Since energy is inversely proportional to wavelength, by the relation E = hcλ, relation (1) will imply,
λ(Kα) > λ(Kβ) < λ(Kγ). (2)
What now about the L and M lines? The diffference between the energy levels decreases as the principal quantum number n increases. This will mean -
(EM−EN) < (EL−EM) < (EK−EL)
→ E(Mα)<E(Lα)<E(Kα) (3)
→ λ(Mα) > λ(Lα) < λ(Kα) (4)
Since energy and wavelength share a reciprocal relation.
Question 13
Figure shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for the tube A in which the potential difference between the target and the filament is VA and the atomic number of the target material is ZA. These quantities are VB and ZB for the other tube. Then,
SOLUTION
Solution : B
In the hint given we asked you to think about two questions:
1) How minimum wavelength is affected by accelerating potential and atomic number?
2) How wavelength of characteristic Kα x-ray is affected by accelerating potential?
Let's see whether answering these two leads us to our solution or not ?
Let's answer the 1st question first.
Minimum wavelength is given by λ0 = hceV0.
We see that it only depends on the accelerating potential but not the atomic number.
Also, a lesser λ0 implies to a higher accelerating potential, hence
VA > VB.
Now let's answer the 2nd question.
Kα refers to the transition of electron from L shell to K . In the process energy is released. If the Z is greater, then the force of attraction for this electron will be higher hence we can conclude that more energy will be released.
It’s also evident from the Moseley's law using which we see that the wavelength of Kα x - ray depends on atomic number but is not related to accelerating potential.
The relation is given by:
√v=a(Z−b)OR√cλ=a(Z−b)
We find that a lesser Kα Characteristic x-ray wavelength implies a higher atomic number.
From the graph;
ZA < ZB.
Question 14
While playing with the Crook’s tube, what was Mr. Roentgen looking for ?
Discovery of Benzene
Behaviour of radiations being generated inside Crook’s tube
Speed of light
Discovery of fundamental particles
SOLUTION
Solution : B
Sneaking through history, we understand that Mr. Roentgen was looking for the behaviour of the radiation being generated inside the Crook’s tube.
Question 15
The wavelength of K-alpha characteristic x-rays only depends upon:
The atomic energy levels
Current in the Coolidge tube setup
Accelerating potential in the Coolidge tube
Can't comment
SOLUTION
Solution : A
Atomic energy levels decide the amount of energy released in an electronic transition, and the wavelength of a photon is dependent on their energy.