If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is
KEY CONCEPT : The energy of nth orbit of hydrogen is given by
Therefore the energy required to remove electron from n = 2 is + 3.4 eV.
At absolute zero, Si acts as
Pure silicon, at absolute zer o, will contain al l the electrons in bounded state. The conduction band will be empty. So there will be no free electrons (in conduction band) and holes (in valence band) due to thermal agitation. Pure silicon will act as insulator.
At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit
(i) electrons
(ii) protons
(iii) He^{2+ }
(iv) neutrons
The emission at instant can be
Charged particles are deflected in magnetic field.
Sodium and copper have work functions 2.3 eV and 4.5 eV respectively. Then the ratio of the wavelengths is nearest to
We know that work function is the energy required and energy E = hu
Formation of covalent bonds in compounds exhibits
For mation of covalent bond is best explain ed by molecular orbital theory.
If N_{0} is the original mass of the substance of halflife period t_{1/2} = 5 years, then the amount of substance left after 15 years is
After every halflife, the mass of the substance reduces to half its initial value.
By increasing the temperature, the specific resistance of a conductor and a semiconductor
Specific resistance is resistivity which is given by
where n = no. of free electrons per unit volume and τ = average relaxation time For a conductor with rise in temperature n increases and τ decreases. But decrease in t is more dominant than increase in n resulting an increase in the value of ρ.
For a semiconductor with rise in temperature, n increases and τ decreases. But the increase in n is more dominant than decrease in τ resulting in a decrease in the value of ρ.
The energy band gap is maximum in
The energy band gap is maximum in insulators.
The part of a transistor which is most heavily doped to produce large number of majority carriers is
Emitter sends the majority charge carrriers towards the collector. Therefore emitter is most heavily doped.
Which of the following are not electromagnetic waves?
β rays are fast moving beam of electrons.
A strip of copper and another of germanium are cooled from room temperature to 80K. The resistance of
The resist ance of metal (like Cu) decr eases with decrease in temperature whereas the resistance of a semiconductor (like Ge) increases with decrease in temperature.
Which of the following radiations has the least wavelength ?
The electromagnetic spectrum is as follows
∴ γrays has least wavelength
When a U^{238} nucleus originally at rest, decays by emitting an alpha particle having a speed ‘u’, the recoil speed of the residual nucleus is
(c) : Linear momentum is conserved
The difference in the variation of resistance with temeperature in a metal and a semiconductor arises essentially due to the difference in the
When the temperature increases, certain boun ded electrons become free which tend to promote conductivity. Simultaneously number of collisions between electrons and positive kernels increases
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is
A nucleus with Z = 92 emits the following in a sequence:
Then Z of the resulting nucleus is
The number of α  particles released = 8 Therefore the atomic number should decrease by 16
The number of β^{} particles released = 4
Therefore the atomic number should increase by 4.
Also the number of β^{+} particles released is 2, which should decrease the atomic number by 2.
Therefore the final atomic number is
92 –16 + 4 – 2 = 78
Two identical photocathodes receive light of frequencies f_{1} and f_{2}. If the velocites of the photo electrons (of mass m) coming out are respectively v_{1} and v_{2}, then
For one photocathode
For another photo cathode
Subtracting (ii) from (i) we get
Which of the following cannot be emitted by radioactive substances during their decay ?
The radioactive substances emit α particles (Helium nucleus), β – particles (electrons) and neutrinoes.
In the nuclear fusion reaction
given that the repulsive potential energy between the two nuclei is ~ 7.7 × 10 ^{14} J , the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann’s Constant k = 1.38 × 10 ^{23} J/K]
The average kinetic energy per molecule
This kinetic energy should be able to provide the repulsive potential energy
Which of the following atoms has the lowest ionization potential ?
The ionisation potential increases from left to right in a period and decreases from top to bottom in a group.
Therefore ceasium will have the lowest ionisation potential.
The wavelengths involved in the spectrum of deuterium are slightly different from that of hydrogen spectrum, because
The wavelength of spectrum is given by
where m = mass of electron
M = mass of nucleus.
For different M, R is different and therefore λ is different
In the middle of the depletion layer of a reverse biased pn junction, the
As in reverse bias, the current through the 0000 is zero through the electric field is also zero.
If the binding energy of the electron in a hydrogen atom is 13.6eV, the energy required to remove the electron from the first excited state of Li ^{++} is
For lithium ion Z = 3 ; n = 2 (for first excited state)
A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
Momentum of photon = E/c
Change in momentum = 2E/c
= momentum transferred to the surface(the photon will reflect with same magnitude of momentum in opposite direction)
According to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal Vs the frequency, of the incident radiation gives as straight the whose slope
From E quation K .E = hvφ
slope of graph of K.E & v is h (Plank's constant) which is same for all metals
The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately.
For the longest wavelength to emit photo electron
A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of 2 : 1. The ratio of their nuclear sizes will be
From conservation of momentum m_{1}v_{1} = m_{2}v_{2}
The binding energy per nucleon of deuteron and helium nucleus is 1.1 MeV and 7 MeV respectively.. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
The nuclear reaction of process is He Energy released = 4 × (7) – 4(1.1) = 23.6 MeV
An αparticle of energy 5 MeV is scattered through 180º by a fixed uranium nucleus. The distance of closest approach is of the order of
KEY CONCEPT :
Distance of closest approach
When npn transistor is used as an amplifier
Electrons move from base to emmitter.
For a transistor amplifier in common emitter configuration for load impedance of 1k Ω (h _{fe} = 50 and h_{oe} = 25) the current gain is
In common emitter configuration current gain
A piece of copper and another of germanium are cooled from room temperature to 77K, the resistance of
Copper is a conductor, so its resistance decreases on decreasing temperature as thermal agitation decreases,; whereas germanium is semiconductor therefore on decreasing temperature resistance increases.
The manifestation of band structure in solids is due to
Pauli’s exclusion principle.
When pn junction diode is forward biased then
Both the depletion region and barrier height is reduced.
If radius of the nucleus is estimated to be 3.6 fermi then the radius of nucleus be nearly
KEY CONCEPT : R = R_{0}(A)^{1/3}
Starting with a sample of pure of it decays into Zn in 15 minutes. The corresponding half life is
7/8 of Cu decays in 15 minutes.
⇒ T = half life period = 15/3 = 5 minutes
A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 1/2 m away,, the number of electrons emitted by photocathode would
When intensity becomes 4 times, no. of photoelectrons emitted would increase by 4 times, since number of electrons emitted per second is directly proportional to intensity.
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV) for the semiconductor is
Band gap = energy of photon of wavelength 2480 nm.
So,
The intensity of gamma radiation from a given source is I.On passing through 36 mm of lead, it is reduced to 1/8. The thickness of lead which will reduce the intensity to I/2 will be
KEY CONCEPT : Intensity I = I_{0} .e ^{μd} ,
Applying logarithm on both sides,
Dividing (i) by (ii),
In a common base amplifier, the phase difference between the input signal voltage and output voltage is
Zero; In common base amplifier circuit, input and output voltage are in the same phase.
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
E will be maximum for the transition for which is maximum. Here n_{2} is the higher energy
level.
Clearly, is maximum for the thirdtran sition, i.e. 2 → 1 . I transition represents the absorption of energy.
If the kinetic energy of a free electron doubles, it’s deBroglie wavelength changes by the factor
deBroglie wavelength,
If K.E is doubled, wavelength becomes
A nuclear transformation is denoted by X (n, α) Which of the following is the nucleus of element X ?
On comparison,
A = 7 + 4 – 1 = 10, z = 3 + 2 – 0 = 5
It is boron _{5}B^{10}
In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be
Input frequency, f = 50 Hz
For full wave rectifier,
In a common base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (β) will be
The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in
This range lies in ultra violet range.
An alph a nucleus of energy bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
Work done to stop the a particle is equal to K.E.
The time taken by a photoelectron to come out after the photon strikes is approximately
The order of time is 10^{–10}5.
When _{3}Li^{7} nuclei are bombar ded by pr oton s, and th e resultant nuclei are _{4}Be^{8}, the emitted particles will be
The energy spectrum of βparticles [number N(E) as a function of βenergy E] emitted from a radioactive source is
The range of energy of βparticles is from zero to some maximum value.
A solid which is not transparent to visible light and whose conductivity increases with temperature is formed by
Van der Waal's bonding is attributed to the attractive forces between molecules of a liquid. The conductivity of semiconductors (covalent bonding) and insulators (ionic bonding) increases with increase in temperature while that of metals (metallic bonding) decreases.
If the ratio of the concentration of electrons to that of holes in a semiconductor is 7/5 and the ratio of currents is 7/4, then what is the ratio of their drift velocities?
The circuit has two oppositively connected ideal diodes in parallel. What is the current flowing in the circuit?
D_{2} is forward biased whereas D_{1} is reversed biased.
So effective resistance of the circuit
R = 4 + 2 = 6Ω
In the following, which one of the diodes reverse biased?
pside con nected to low potential and nside is connected to high potential.
The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows
As λ decreases, n increases and hence the speed of photoelectron increases. The chances of photo electron to meet the anode increases and hence photo electric current increases.
If the binding energy per nucleon in and nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction energy of proton must be
Let E be the energy of proton, then
E + 7 × 5.6 = 2 × [4 × 7.06]
⇒ E = 56.48  39.2= 17.28MeV
The 'rad' is the correct unit used to report the measurement of
The risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue.
Absorbed dose is expressed in rad. A rad is equal to 100 ergs of energy absorbed by 1 gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist L. H. Gray); one gray equals 100 rad.
If the lattice constant of this semiconductor is decreased, then which of the following is correct?
A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice. The spacing between unit cells in various directions is called its lattice parameters or constants. Increasing these lattice constants will increase or widen the bandgap (E_{g}), which means more energy would be required by electrons to reach the conduction band from the valence band. Automatically E_{c} and E_{v} decreases.
The rms value of the electric field of the light coming from the Sun is 720 N/C. The average total energy density of the electromagnetic wave is
E_{rms }= 720
The average total energy density
If M_{O} is the mass of an oxygen isotope _{8}O^{17 },M_{P} and M_{N} are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is
Binding energy
= [ZM_{P} + (A – Z)M_{N} – M]c^{2}
= [8M_{P} + (17 – 8)M_{N} – M]c^{2}
= [8M_{P} + 9MN – M]c^{2}
= [8M_{P} + 9M_{N} – Mo]c^{2}
In gamma ray emission from a nucleus
There is no change in the proton number and the neutron number as the gemission takes place as a result of excitation or deexcitation of nuclei. γrays have no charge or mass.
If in a pn junction diode, a square input signal of 10 V is applied as shown
Then the output signal across R_{L} will be
The current will flow through R_{L} when the diode is forward biased.
Photon of frequency v has a momentum associated with it. If c is the velocity of light, the momentum is
Energy of a photon of frequency n is given by E = hv .
The halflife period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially they have the same number of atoms. Then
According to question,
Now, the rate of decay is given by
∴ Y will decay faster than X. [∵N is some]
Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate ?
Si and Ge are semiconductors but C is an insulator.
Also, the conductivity of Si and Ge is more than C because the valence electrons of Si, Ge and C lie in third, fourth and second orbit respectively.
Which of the following transitions in hydrogen atoms emit photons of highest frequency?
We have to find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options (c) and (d).
Frequency is given by
For transition from n = 6 to n = 2,
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).
Q. Electrons accelerated by potential V are diffracted from a crystal. If d = 1Å and i = 30°, V should be about (h = 6.6 × 10 ^{– 34} Js, me = 9.1 × 10^{–31} kg, e = 1.6 × 10 ^{– 19} C)
Using Bragg's equation 2d sinθ = nλ
Here n = 1, θ = 90 – i = 90 – 30 = 60°
∴ 2d sin θ = λ ......(i)
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure).
Q. If a strong diffraction peak is observed when electrons are incident at an angle ‘i’ from the normal to the crystal planes with distance ‘d’ between them (see figure), de Broglie wavelength λ_{dB} of electrons can be calculated by the relationship (n is an integer)
2d cos i = nλ_{dB}
This question contains Statement1 and statement2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and
Statement2 : For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z.
We know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Therefore statement (1) is correct.
The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing Z and for light nuclei, B.E/nucleon increases with increasing Z.
A working transistor with its three legs marked P, Q and R is tested using a multimeter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multimeter. Which of the following is true for the transistor?
It is a npn transistor with R as base.
Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the n^{th} orbital of the electron is found to be ‘r_{n}’ and the kinetic energy of the electron to be ‘T_{n}’.
Then which of the following is true
When
⇒ mv^{2} = constat ⇒ kinetic energy is constant
⇒ T is independent of n.
In the circuit below, A and B represent two inputs and C represents the output.
The circuit represents
The truth table for the above logic gate is : This truth table follows the boolean algebra C = A + B which is for OR gate
The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from :
It is given that transition from the state n = 4 to n = 3 in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is 5 → 4 .
The surface of a metal is illuminted with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is :
(h_{c} = 1240 eV.nm)
The above is a plot of binding energy per nucleon E_{b}, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions :
(i) A + B → C + ε
(ii) C → A + B + ε
(iii) D + E → F + ε and
(iv) F→ D + E + ε,
where ε is the energy released? In which reactions is ε positive?
For A + B → C + ε,ε is positive. This is because E_{b} for C is greater than the E_{b} for A and B.
Ag ai n for F → D +E + ε,ε is positive. Th is is because E_{b} for D and E is greater than E_{b} for F .
The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. Pick out the correct output waveform.
Output is
Here Thus it is an AND gate for which truth table is
A pn junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit.
The current (I) in the resistor (R) can be shown by :
We know that a single pn junction diode connected to an ac source acts as a half wave rectifier [Forward biased in one half cycle and reverse biased in the other half cycle].
Statement 1 : When ultraviolet light is inciden t on a photocell, its stopping potential is V_{0} and the maximum kinetic energy of the photoelectrons is K_{max} .When the ultraviolet light is replaced by Xrays, both V_{0} and K_{max} increase.
Statement 2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
We know that
where, φ is the work function .
Hence, as v increases (note that frequency of Xrays is greater than that of U.V. rays), both V_{0} and K_{max} increase. So statement  1 is correct
A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.
Q. The binding energy per nucleon for the parent nucleus is E_{1} and that for the daughter nuclei is E_{2}. Then
In nuclear fission, the binding energy per nucleon of daughter nuclei is greater than the parent nucleus.
A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.
Q. The speed of daughter nuclei is
By conservation of energy,
where v is the speed of the daughter nuclei
A radioactive nucleus (initial mass number A and atomic number Z emits 3 a  particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be
The combination of gates shown below yields
The final boolean expression is,
If a source of power 4kW produces 10^{20} photons/second , the radiation belongs to a part of the spectrum called
This question has Statement – 1 and Statement – 2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement – 1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals.
Statement – 2 : The state of ionosphere varies from hour to hour, day to day and season to season.
For long distance communication, sky wave signals are used.
Also, the state of ionosphere varies every time. So, both statements are correct.
Energy required for the electron excitation in Li^{++} from the first to the third Bohr orbit is :
Energy of excitation,
The half life of a radioactive substance is 20 minutes. The approximate time interval (t_{2} – t_{1}) between the time t_{2} when 2/3 of it had decayed and time t_{1} when 1/3 of it had decayed is :
Number of undecayed atom after time t_{2} ;
Number of undecayed atom after time t_{1};
Solving (iii) and (iv), we get t_{2} – t_{1} = 20 min
This question has Statement – 1 and Statement – 2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement – 1: A metallic surface is irradiated by a monochromatic light of frequency v > v_{0} (the threshold frequency). The maximum kinetic energy and the stopping potential are K_{max} and V_{0} respectively. If the frequency incident on the surface is doubled, both the K_{max} and V_{0} are also doubled.
Statement – 2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surfac
By Einstein photoelectric equation,
K_{max} = eV_{0} = hv – hv_{0}
When v is doubled, K_{max} and V_{0} become more than double.
Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be :
The possible number of the spectral lines is given
Truth table for system of four NAND gates as shown in figure is :
By expanding this Boolen expression
Thus the truth table for this expression should be (1).
A radar has a power of 1kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500 m.
The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 × 10^{6}m) is :
Let d is the maximum distance, upto it the objects From ΔAOC
Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron= 1.6725 × 10^{–27} kg, mass of proton = 1.6725 × 10^{–27} kg,mass of electron = 9 × 10^{–31} kg).
The mass defect during the process
= 1.6725 × 10^{–27} – (1.6725 × 10^{–27}+ 9 × 10^{–31}kg) = – 9 × 10^{–31} kg
The energy released during the process
E = Δmc^{2}
E = 9 × 10^{–31}× 9 × 10^{16} = 81 × 10^{–15} Joules
A diatomic molecule is made of two masses m_{1} and m_{2} which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by : (n is an integer)
The energy of the system of two atoms of diatomic molecule
where I = moment of inertia
ω = Angular velocity = L/I,
L = Angular momentum
A diode detector is used to detect an amplitudemodulated wave of 60% modulation by using a condenser of capacity 250 picofarad in parallel with a load resistance 100 kilo ohm.
Find the maximum modulated frequency which could be detected by it.
Given : Resistance R = 100 kilo ohm = 100 × 10 ^{3} ω Capacitance C = 250 picofarad = 250 × 10^{–12}F
= 2.5 × 10^{–5} sec
The higher frequency whcih can be detected with tolerable distortion is
This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated singnal voltage for proper detection of mdoulated signal.
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :
From question,
B0 = 20 nT = 20 × 10^{–9}T
(∴ velocity of light in vacuum C = 3 × 10^{8 }ms^{–1})
= 6 V/m.
The anode voltage of a photocell is kept fixed. Th e wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows :
As λ is increased, there will be a value of λ above which photoelectrons will be cease to come out so photocurrent will become zero. Hence (d) is correct answer.
The IV characteristic of an LED is
For same value of current higher value of voltage is required for higher frequency hence (1) is correct answer.
In a hydrogen like atom electron make transition from an energy level with quantum number n to another with quantum number (n – 1). If n>>1, the frequency of radiation emitted is proportional to :
Th e cur r en t voltage r elation of a diode is given by , where the applied voltage V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ±0.01V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?
The current voltage relation of diode is
(By exponential function)
During the propagation of electromagnetic waves in a medium:
Electric energy density
Magnetic energy density
Energy is equally divided between electric and magnetic field
The radiation corresponding to 3→2 transition of hydrogenatom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10^{–4} T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to:
Radius of circular path followed by electron is given by,
For transition between 3 to 2.
Work function = 1.88 eV – 0.8 eV = 1.08 eV ≈ 1.1eV
Hydrogen Deuterium singly ionised Helium and doubly ionised lithium all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are respectively then approximately which one of the following is correct?
Wave number
By question n = 1 and n_{1} = 2
The forward biased diode connection is:
For forward bias, pside must be at higher potential than nside. ΔV = (+)Ve
Match List  I (Electromagnetic wave type) with List  II (Its association/application) and select the correct option from the choices given below the lists:
List 1 List 2
1. Infrared waves (i) To treat muscular strain
2. Radio waves (ii) For broadcasting
3. Xrays (iii) To detect fracture of bones
4. Ultraviolet rays (iv) Absorbed by the ozone layer of the atmosphere
1 2 3 4
(1) Infrared rays are used to treat muscular strain because these are heat rays. (2) Radio waves are used for broadcasting because these waves have very long wavelength ranging from few centimeters to few hundred kilometers (3) Xrays are used to detect fracture of bones because they have high penetrating power but they can't penetrate through denser medium like bones. (4) Ult r aviol et r a ys ar e absorbed by ozone of the atmosphere.
A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is :
∴E_{0} = √6 = 2.45V/m
A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are :
Amplitude modulated wave consists of three frequencies are
i.e. 2005 KHz, 2000KHz, 1995 KHz
As an electron makes a transition from an excited state to the ground state of a hydrogen  like atom/ion :
Match List  I (Fundamental Experiment) with List  II (its conclusion) and select the correct option from the choices given below the list:
FrankHertz experiment  Discrete energy levels of atom Photoelectric effect  Particle nature of light Davison  Germer experiment  wave nature of electron.
For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and β is :
We know that
Also I_{e} = I_{b }+ I_{c}
Option (b) and (d) are therefore correct.
If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is :
In case of an 'OR' gate the input is zero when all inputs are zero. If any one input is ' 1', then the output is '1'.
Choose the correct statement :
In amplitude modulation, the amplitude of the high frequency carrier wave made to vary in proportional to the amplitude of audio signal.
Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to the speed of the fastest emitted electronwill be:
Halflives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed number of A and B nuclei will be :
For A_{t½} = 20 min, t = 80 min, number of half lifes n = 4
∴ Nuclei remaining Therefore nuclei decayed
For B_{t½} = 40 min., t = 80 min, number of half lifes n = 2
∴ Nuclei remaining =N_{0}/2^{2 }.Therefore nuclei decayed
∴ Required ratio
Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d) :
Graph (a) is for a simple diode.
Graph (b) is showing the V Break down used for zener diode.
Graph (c) is for solar cell which shows cutoff voltage and open circuit current.
Graph (d) shows the variation of resistance h and hence current with intensity of light.
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