The plate resistance of a triode is 3 × 10^{3} ohms and its mutual conductance is 1.5 × 10^{–3} amp/volt. The amplification factor of the triode is
KEY CONCEPT : We know that µ = g_{m} × r_{0}
where µ = amplification factor,
gm = mutual conductance
r_{0} = plate resistance
∴ µ = 3 × 10^{3} × 1.5 × 10^{–3} = 4.5
The half life of radioactive Radon is 3.8 days. The time at the end of which 1/20 th of the radon sample will remainundecayed is (given log_{10 }e = 0.4343)
t_{1/2} = 3.8 day
If the initial number of atom is a = A_{0} then after time t the number of atoms is a/20 = A. We have to find t.
An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of
One point charge is uranium nucleus
The other point charge is α particle ∴ q_{2 }= + 2_{e}
Here the loss in K.E. = Gain in P.E. (till αparticle reaches the distance d)
= 529.92 × 10^{–16} m
= 529.92 × 10^{–14} cm
= 5.2992 × 10^{–12} cm
Beta rays emitted by a radioactive material are
βparticles are charged particles emitted by the nucleus.
If elements with principal quantum number n > 4 were not allowed in nature, the number of possible elements would be
KEY CONCEPT : The maximum number of electrons in an orbit is 2n^{2}. n > 4 is not allowed.
Therefore the number of maximum electron that can be in first four orbits are 2 (1)^{2} + 2 (2)^{2} + 2 (3)^{2} + 2 (4)^{2}
= 2 + 8 + 18 + 32 = 60
Therefore, possible element are 60.
Consider the spectral line resulting from the transition n = 2 → n = 1 in the atoms and ions given below. The shortest wavelength is produced by
We know that
λ is shortest when 1/λ is largest i.e., when Z has a highervalue. Z is highest for lithium.
The equation represents
Fast neutrons can easily be slowed down by
Fast neutrons can be easily slowed down by passing them through water.
Consider α particles, β particles and γ  rays, each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are:
Note : The penetrating power is dependent on velocity.
For a given energy, the velocity of γ radiation is highest and αparticle is least.
An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy in (eV) required to remove both the electrons from a neutral helium atom is
When one e^{–} is removed from neutral helium atom, it becomes a one e^{–} species.
For one e^{–} species we know
For helium ion, Z = 2 and for first orbit n = 1.
∴ Energy required to remove this e^{–} = + 54.4 eV
∴ Total energy required = 54.4 + 24.6 = 79 eV
A radioactive material decays by simultaneous emission of two particles with respective halflives 1620 and 810 years.
The time, in years, after which onefourth of the material remains is
when N_{o} is initial number of atoms
The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature
KEY CONCEPT : For a semi conductor n = n_{0}e^{–Eg/kT} where n_{0} = no. of free electrons at absolute zero, n = no. of free electrons at T kelvin, E_{g} = Energy gap, k = Boltzmann constant.
As E_{g} increases, n decreases exponentially.
A fullwave rectifier circuit along with the output is shown in Figure. The contribution (s) from the diode 1 is (are)
As shown in the fig. (i) during one half cycle the polarity of P and S are opposite such that diode (1) is reversed biased and hence non conducting.
During the other half cycle, diode (1) gets forward biased and is conducting. Thus diode (1) conducts in one half cycle and does not conduct in the other so the correct option is (b) (a and c.)
As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is
KEY CONCEPT :
Therefore, ground state energy of doubly ionized lithium atom (Z = 3, n = 1) will be
∴ Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4eV.
The circuit shown in the figure contains two diodes each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6V, the current through the 100 ohm resistance (in amperes) is
In the circuit, diode D_{1} is forward biased, while D_{2} is reverse biased. Therefore, current i (through D_{1} and 100Ω resistance) will be
Here, 50Ω is the resistance of D_{1} in forward biasing.
Which of the following statements is not true?
In ntype semiconductors, electrons are the majority charge carriers.
The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential, in volt, is
Stopping potential is the negative potential applied to stop the electrons having maximum kinetic energy.
Therefore, stopping potential will be 4 volt.
In hydrogen spectrum the wavelength of H_{α} line is 656 nm, whereas in the spectrum of a distant galaxy, H_{α} line wavelength is 706 nm. Estimated speed of the galaxy with respect to earth is,
KEY CONCEPT : According to Doppler’s effect of light, the wavelength shift is given by
A particle of mass M at rest decays into two particles of masses m_{1} and m_{2}, having nonzero velocities. The ratio of the de Broglie wavelengths of the particles, λ_{1}/λ_{2}, is
Applying conservation of linear momentum, Initial momentum = Final momentum
0 = m_{1}v_{1} – m_{2}v_{2} ⇒ m_{1}v_{1} = m_{2}v_{2}
Which of the following is a correct statement?
Beta rays are same as cathode rays as both are stream of electrons.
Order of magnitude of density of uranium nucleus is, [m_{p} = 1.67 × 10^{–27}kg]
Nuclear density of an atom of mass number A,
^{22}Ne nucleus, after absorbing energy, decays into two aparticles and an unknown nucleus. The unknown nucleus is
The new element X has atomic number 6. Therefore, it is carbon atom.
Binding energy per nucleon vs mass number curve for nuclei is shown in the Figure. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is
KEY CONCEPT : Energy is released when stability increases. This will happen when binding energy per nucleon increases.
Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength λ (given in terms of the Rydberg constant R for the hydrogen atom) equal to
KEY CONCEPT :
For ordinary hydrogen atom, longest wavelength
With hypothetical particle, required wavelength
The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true ?
NOTE : As the electron comes nearer to the nucleus the potential energy decreases
Two radioactive materials X_{1} and X_{2} have decay constants 10λ and λ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X_{1} to that of X_{2} will be 1/e after a time
Electrons with energy 80 keV are incident on the tungsten target of an Xray tube. Kshell electrons of tungsten have 72.5 keV energy. Xrays emitted by the tube contain only
KEY CONCEPT :
Energy of incident electrons is greater than the ionization energy of electrons in Kshell, the Kshell electrons will be knocked off. Hence, characteristic Xray spectrum will be obtained.
The electron emitted in beta radiation originates from
Note : In a nucleus neutron converts into proton as follows n → p^{+} + e^{–1} Thus, decay of neutron is responsible for bradiation origination
The transition from the state n = 4 to n = 3 in a hydrogenlike atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition
For 2 to 1, 3 to 2 and 4 to 2 we get energy that n = 4 to n = 3,
I.R. radiation has less energy than U.V. radiation.
The intensity of Xrays from a Coolidge tube is plotted against wavelength λ as shown in the figure. The minimum wavelength found is λ_{C} and the wavelength of the K_{α} line is λ_{K}. As the accelerating voltage is increased
KEY CONCEPT :
In case of Coolidge tube
Thus the cut off wavelength is inversely proportional to accelerating voltage. As V increases, λ_{c} decreases. λ_{k} is the wavelength of K_{∝} line which is a characteristic of an atom and does not depend on accelerating voltage of bombarding electron since λ_{k} always refers to a photon wavelength of transition of e^{–} from the target element from 2 → 1.
The above two facts lead to the conclusion that λ_{k} – λ_{c} increases as accelerating voltage is increased.
A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life time of one species is τ and that of the other is 5τ. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figures best represent the form of this plot?
....(i)
Adding (i) and (ii) we get
N = N_{1} + N_{2} = N_{0} (e ^{t/τ}+e^{t/5τ})
(a) is NOT the correct option as there is a time τ for which N is constant which means for time τ there is no process of radioactivity which does not makes sense. (b) and (c) shows intermediate increase in the number of radioactive atom which is IMPOSSIBLE as N will only decrease exponentially.
The potential difference applied to an Xray tube is 5kV and the current through it is 3.2mA. Then the number of electrons striking the target per second is
KEY CONCEPT :
No. of electrons striking the target per second
A Hydrogen atom and a Li^{++ }ion are both in the second excited state. If ℓ _{H} and ℓ _{Li} are their respective electronic angular momenta, and E_{H} and E_{Li} their respective energies, then
The halflife of ^{215}At is 100 μ s. The time taken for the radioactivity of a sample of ^{215}At to decay to 1/16^{th} of its initial value is
A = A_{0} (1/2)^{n}; n = number of half lives.
Which of the following processes represents a γ decay ?
In γdecay, the atomic number and mass number do not change.
The electric potential between a proton and an electron is given by where r_{0} is a constant. AssumingBohr’s model to be applicable, write variation of r_{n} with n, n being the principal quantum number?
Given potential energy between electron and proton
But this force acts as centripetal force
By Bohr’s postulate, mvr = nh/2π ...(ii)
From (i) and (ii),
If the atom _{100}Fm^{257} follows the Bohr model and the radius of _{100}Fm^{257} is n times the Bohr radius, then find n.
KEY CONCEPT : For an atom followig Bohr ’s model, the radius is given by
= Bohr’s radius and m = orbit number.
For Fm, m = 5 (Fifth orbit in which the outermost electron is present)
For uranium nucleus how does its mass vary with volume?
KEY CONCEPT : We know that radius of the nucleus R = R_{0}A^{1/3}, where A is the mass number.
Volume ∝ mass.
A nucleus with mass number 220 initially at rest emits an α particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the α particle
By conservation of momentum, p_{1} = p_{2}
Solve equation (i) and (ii)
In a photoelectric experiment anode potential is plotted against plate current.
From the graph it is clear that A and B have the same stopping potential and therefore, the same frequency.
Also, B and C have the same intensity.
A 280 days old radioactive substance shows an activity of 6000 dps, 140 days later its activity becomes 3000 dps.
What was its initial activity?
In two half lives, the activity will remain 1/4 of its initial activity.
A pr oton has kinetic ener gy E = 100 keV which is equal to that of a photon. The wavelength of photon is λ_{2} and that of proton is λ_{1}. The ration of λ_{2}/λ_{1} is proportional to
For photon,
For proton,
...(ii)
K_{α} wavelength emitted by an atom of atomic number Z = 11 is λ. Find the atomic number for an atom that emits K_{α} radiation with wavelength 4λ.
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV.
After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector?
Initially a photon of energy 10.2eV collides inelastically with a hydrogen atom in ground state. For hydrogen atom,
The electron of hydrogen atom will jump to second orbit after absorbing the photon of energy 10.2 eV. The electron jumps back to its original state in less than microsecond and release a photon of energy 10.2 eV.
Another photon of energy 15 eV strikes the hydrogen atom inelastically. This energy is sufficient to knock out the electron from the atom as ionisation energy is 13.6 eV. The remaining energy of 1.4 eV is left with electron as its kinetic energy.
A beam of electron is used in an YDSE experiment. The slit width is d. When the velocity of electron is increased, then
Note : Since electron shows wave nature, it will show the phenomenon of interference.
When speed of electron increases, λ will decrease. The distance between two consecutive fringes
As λ decreases, β also decrease.
If a star can convert all the He nuclei completely into oxygen nuclei, the energy released per oxygen nuclei is [Mass of He nucleus is 4.0026 amu and mass of Oxygen nucleus is 15.9994 amu]
B.E. = Δm × 931.5 MeV
= (4 × 4.0026 – 15.9994) × 931.5 = 10.24 MeV
is a radioactive substance having half life of 4 days. Find the probability that a nucleus undergoes decay after two half lives
For a nucleus to disintegr ate in two half life, the probability is 3/4 as 75% of the nuclei will disintegrate in this time.
In the options given below, let E denote the rest mass energy of a nucleus and n a neutron.The correct option is
Iodine and Yttrium are medium sized nuclei and therefore, have more binding energy per nucleon as compared to Uranium which has a big nuclei and less B.E./nucleon.
In other words, Iodine and Yttrium are more stable and therefore possess less energy and less rest mass. Also when Uranium nuclei explodes, it will convert into I and Y nuclei having kinetic energies.
The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is
The smallest frequency and largest wavelength in ultraviolet region will be for transition of electron from orbit 2 to orbit 1.
The highest frequency and smallest wavelength for infrared region will be for transition of electron from ∞ to 3rd orbit.
Electrons with deBroglie wavelength λ fall on the target in an Xray tube. The cutoff wavelength of the emitted Xrays is
The cut off wavelength is given by
...(i)
According to de Broglie equation
From (i) and (ii),
Which one of the following statements is WRONG in the context of Xrays generated from a Xray tube?
The continuous spectrum depends on the accelerating voltage. It has a definite minimum wavelength.
Greater the accelerating voltage for electrons, higher will be the kinetic energy it attains before striking the target, higher will be the frequency of X  rays and smaller will be the wavelength. The wavelength of continuous X  rays is independent of the atomic number of target material.
A radioactive sample S_{1} having an activity 5_{µ}Ci has twice the number of nuclei as another sample S_{2} which has an activity of 10 _{µ}Ci. The half lives of S_{1} and S_{2} can be
Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions φ_{p} = 2.0 eV, φ_{q} = 2.5 eV and φ_{r} = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct IV graph for the experiment is [Take hc = 1240 eV nm]
The energy possessed by photons of wavelen gth 550 nm is 1240/550 = 2.25eV
The energy possessed by photons of wavelength 450 nm is 1240/450 = 2.76eV
The energy possessed by photons of wavelength 350 nm is 1240/350 = 3.5.eV
For metal plate p :
φ_{p} = 2 eV.
Photons of wavelength 550 nm will not be able to eject electrons and therefore, the current is smaller than p.
The work function is greater than q therefore the stopping potential is lower in comparison to p.
For metal plate r :
φ_{r} = 3 eV
Only wavelength of 350 nm will be able to eject electrons and therefore, current is minimum. Also the stopping potential is least.
The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A°. The wavelength of the second spectral line in the Balmer series of singlyionized helium atom is
The wave length of first spectral line in the Balmer series of hydrogen atom is 6561Å . Here n_{2} = 3 and n_{1} = 2
For the second spectral line in the Balmer series of singly ionised helium ion n_{2} = 4 and n_{1} = 2 ; Z = 2
Dividing equation (i) and equation (ii) we get
A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3×10^{8} ms^{–1}. The final momentum of the object is
If λ_{Cu} is the wavelength of K_{α} Xray line of copper (atomic number 29) and λ_{Mo} is the wavelength of the K_{α} Xray line of molybdenum (atomic number 42), then the ratio λ_{Cu}/λ_{Mo} is close to
A metal surface is illuminat ed by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u_{1} and u_{2}, respectively. If the ratio u_{1} : u_{2} = 2 : 1 and h_{c} = 1240 eV nm, the work function of the metal is nearly
Dividing the above two equations, we get
In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ) of incident light and the corresponding stopping potential (V_{0}) are given below :
Given that c = 3 × 10^{8}m s^{–1 }and e = 1.6 × 10^{–19} C , Planck's constant (in units of J s) found from such an experiment is
From the first two values given in data
Similarly if we calculate h for the last two values of data h = 6.4 × 10^{–34}Js
The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by
The measured masses of the neutron are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the nuclei are same, 1 u = 931.5 Me V/c^{2} (c is the speed of light) and e^{2}/(4 πε_{0}) = 1.44 MeV fm. Assuming that the difference between the binding energies of is purely due to the electrostatic energy, the radius of either of the nuclei is (1 fm = 10^{–15} m)
Binding energy of nitrogen atom
= [8 × 1.008665 + 7 × 1.007825 – 15.000109] × 931
Binding energy of oxygen atom
= [7 × 1.008665 + 8 × 1.007825 – 15.003065] × 931
∴ Difference = 0.0037960 × 931 MeV ...(I)
From (i) & (ii)
An accident in a nuclear labora tory resulted in deposition of a certain amount of radioactive material of halflife 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?
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