Q.19. The element CuriumCm has a mean life of 10^{1}^{3} seconds.
Its pirmary decay modes are spontaneous fission and adecay, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV of energy. The masses involved in adecay are as follows: = 244.064100 u and. Calculate the power output from a sample of 10^{20} Cm atoms. (1 u = 931 MeV/c^{2}.) (1997  5 Marks)
Ans. 3.32 × 10^{–5} W
Solution. The reaction involved in adecay is
Mass defect
= (248.072220 – 244.064100 – 4.002603) u
= 0.005517u
Therefore, energy released in adecay will be
E_{α} = (0.005517 × 931) MeV = 5.136 MeV
Similarly, E_{fissio}_{n} = 200 MeV (given)
Mean life is given as t_{mean}
∴ Disintegration constant λ = 10^{–13} s^{–1}
Rate of decay at the moment when number of nuclei are 10^{20} is
Of these distintegrations, 8% are in fission and 92% are in adecay.
Therefore, energy released per second
= (0.08 × 10^{7} × 200 + 0.92 × 10^{7} × 5.136) MeV
= 2.074 × 10^{8} MeV
∴ Power output (in watt) = Energy released per second (J/s)
= (2.074 × 10^{8}) (1.6 × 10^{–13})
Q.20. Nuclei of a radioactive element A are being produced at a constant rate α. The element has a decay constant λ. At time t = 0, there are N_{0} nuclei of the element. (1998  8 Marks)
(a) Calculate the number N of nuclei of A at time t.
(b) If α = 2N_{0}λ, calculate the number of nuclei of A after one halflife of A, and also the limiting value of N as t →∞ .
Ans.
Solution.
(a) Let at time ‘t’ number of radioactive nuclei are N.
Net rate of formation of nuclei of A.
Solving this equation, we get
........(1)
(b) Substituting a = 2λN_{0} and
(ii) Substituting α = 2λ N_{0} and t → ∞ in equation (1), we get
Q.21. Photoelectrons are emitted when 40 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing αparticles. A maximum energy electron combines with an α–particle to form a He^{+} ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 to 4 eV range, that are likely to be emitted during and after the combination. [Take h = 4.14×10^{–15} eV.s.] (1999  5 Marks)
Ans. 3.4 eV, 3.84 eV and 2.64 eV
Solution. The energy of the incident photon is
The maximum kinetic energy of the emitted electrons is E_{max} = E_{1} – W = 3.1 eV – 1.9 eV = 1.2eV
It is given that,
The fourth excited state implies that the electron enters in the n = 5 state.
In this state its energy is
The energy of the emitted photon in the above combination reaction is E = E_{max} + (– E_{5}) = 1.2 eV + 2.18 eV = 3.4 eV
Note : After the recombination reaction, the electron may undergo transition from a higher level to a lower level thereby emitting photons.
The energies in the electronic levels of He^{+} are
The possible transitions are n = 5 → n = 4
ΔE = E_{5} – E_{4} = [– 2.18 – (– 3.4)] eV = 1.28 eV
n = 5 → n = 3
ΔE = E_{5}– E_{3} = [– 2.18 – (– 6.04)] eV = 3.84 eV
n = 5 → n = 2
ΔE = E_{5} – E_{2} = [– 2.18 – (– 13.6)] eV = 11.4 eV
n = 4 → n = 3
ΔE = E_{4} – E_{3} = [– 3.4 – (– 6.04)] eV = 2.64 eV
Hence, the photons that are likely to be emitted in the range of 2 eV to 4 eV are 3.4 eV, 3.84 eV and 2.64 eV.
Q.22. A hydrogenlike atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) for this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during deexcitation. Ground state energy of hydrogen atom is –13.6 eV. (2000  6 Marks)
Ans. 2, 4, 10.5 eV
Solution. Energy for an orbit of hydrogen like atoms is
For transition from 2n orbit to 1 orbit
Also for transition 2n → n.
From (i) and (ii)
For minimum energy = Transition from 4 to 3.
Hence n = 2, Z = 4, E_{min} = 10.5 eV
Q.23. When a beam of 10.6 eV photons of intensity 2.0 W/m^{2} falls on a platinum surface of area 1.0 × 10^{4} m^{2} and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons.
Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV). Take 1eV = 1.6 × 10^{19} J. (2000  4 Marks)
Ans. 6.25 × 10^{11}, 0 eV, 5 eV
Solution. No. of photons/sec
No. of photons incident per second
As 0.53% of incident photon can eject photoelectrons
∴ No. of photoelectrons ejected per second
Minimum energy = 0 eV,
Maximum energy = (10.6 – 5.6) eV = 5 eV
Q.24. In a nuclear reaction ^{235}U undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 years, find the total mass of uranium required. (2001  5 Marks)
Ans. 38451 Kg
Solution. The formula for η of power will be
Energy required for this power is given by
E = P × t
= 10^{10} × 86,400 × 365 × 10
= 3.1536 × 10^{18} J
200 × 1.6 × 10–13 J of energy is released by 1 fission
∴ 3.1536 × 10^{18} J of energy is released by
= 0.9855 × 10^{29} fission
= 0.985 × 10^{29} of U^{235} atoms.
6.023 × 10^{23} atoms of Uranium has mass 235g
∴ 0.9855 × 10^{29} atoms of Uranium has
Q.25. A nucleus at rest undergoes a decay emitting an a particle of deBroglie wavelength λ = 5.76 × 10^{15} m. If the mass of the daughter nucleus is 223.610 amu and that of the a particles is 4.002 amu, determine the total kinetic energy in the final state. Hence, obtain the mass of the parent nucleus in amu. (1 amu = 931.470 MeV/c^{2}) (20015 Marks)
Ans. 10^{–12} J, 227.62 amu
Solution. Let the reaction be
Here, m_{y} = 223.61 amu and m_{α} = 4.002 amu
We know that
Applying eq. (i) for Y and α, we get
= 0.0982243 × 10–11 = 0.982 × 10–12J
Similarly (K.E.)Y = 0.0178 × 10^{–12} J
Total energy = 10^{–12} J
We know that E = Δmc^{2}
The mass of the parent nucleus X will be
m_{x} = m_{y} + m_{a} + Δ_{m}
= 223.61 + 4.002 + 0.00665 = 227.62 amu
Q.26. A radioactive nucleus X decays to a nucleus Y with a decay constant λ_{x} = 0.1 s^{1}. Y further decays to a stable nucleus Z with a decay constant λ_{Y} = 1/30 s^{1}. Initially, there are only X nuclei and their number is N_{o} = 10^{20}. Set up the rate equations for the populations of X, Y and Z. The population of Y nucleus as a function of time is given by N_{Y}(t) = (N_{o}λ_{X}/ (λ_{X}  λ_{Y})){exp(λ_{Y}t)exp(λ_{x}t)}. Find the time at which N_{Y} is maximum and determine the populations X and Z at that instant. (20015 Marks)
Ans.
Solution.
The rate of equation for the population of X, Y and Z will be
⇒ On integration, we get
N_{x} = N_{0}e ^{λxt} ...(iv)
Given
To determine the maximum N_{Y}, we find
From (ii)
λ_{y} N_{y} + λ_{x}N_{x} = 0
⇒ λ_{x} N_{x} = λ_{y}N_{y} .......(v)
Q.27. A hydrogenlike atom (descr ibed by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between 0.85 eV and 0.544 eV (including both th ese values). (2002  5 Marks)
(a) Find the atomic number of the atom.
(b) Calculate the smallest wavelength emitted in th ese transitions.
(Take hc = 1240 eVnm, ground state energy of hydrogen atom = –13.6 eV)
Ans. (a) 3 (b) 4052 nm
Solution. If x is the difference in quantum number of the states than ^{x+1}C_{2} = 6 ⇒ x = 3
Solving (i) and (ii) we get n = 12 and z = 3
(b) Smallest wavelength λ is given by
Solving, we get λ = 4052 nm.
Q.28. Two metallic plates A an d B, each of area 5 × 10^{–4} m^{2}, are placed parallel to each other at a separation of 1 cm. Plate B carries a positive charge of 33.7 × 10^{–12} C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 10^{16} photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 106 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2eV. Determine (2002  5 Marks)
(a) the number of photoelectrons emitted up to t = 10 s,
(b) the magnitude of the electric field between the plates A and B at t = 10 s, and
(c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B.
Neglect the time taken by the photoelectron to reach plate B.
Take ε_{0} = 8.85 × 10^{–12} C^{2}/Nm^{2}
Ans. (a) 5 × 10^{7}
(b) 2000 N/C
(c) 23 eV
Solution. Number of electrons falling on the metal plate A = 10^{16} × (5 × 10^{–4})
∴ Number of photoelectrons emitted from metal plate A upto 10 seconds is
(b) Charge on plate B at t = 10 sec
Q_{b} = 33.7 × 10^{–12} – 5 × 10^{7} × 1.6 × 10^{–19} = 25.7 × 10^{–12} C
also Q_{a} = 8 × 10^{–12} C
(c) K.E. of most energetic particles
= (hv –φ) + e (Ed) = 23 eV
Note : (hv – φ) is energy of photoelectrons due to light e (Ed) is the energy of photoelectrons due to work done by photoelectrons between the plates.
Q.29. Frequency of a photon emitted due to transition of electron of a certain element from L to K shell is found to be 4.2 × 10^{18} Hz. Using Moseley’s law, find the atomic number of the element, given that the Rydberg’s constant R = 1.1 × 10^{7} m^{–1}. (2003  2 Marks)
Ans. 42
Solution. According to Bohr ’s model, the energy released during transition from n_{2} to n_{1} is given by
For transition from L shell to K shell
b = 1, n_{2} = 2, n_{1} = 1
On putting the value of R = 1.1 × 10^{7} m^{–1} (given),
c = 3 × 10^{8} m/s, we get Z = 42
Q.30. A radioactive sample emits n βparticles in 2 sec. In next 2 sec it emits 0.75 n βparticle, what is the mean life of the sample? (2003  2 Marks)
Ans.
Solution.
Q.31. In a photoelectric experiment set up, photons of energy 5 eV falls on the cathode having work function 3 eV. (a) If the saturation current is i_{A} = 4μA for intensity 10^{–5} W/m^{2}, then plot a graph between anode potential and current. (b) Also draw a graph for intensity of incident radiation 2 × 10^{–5} W/m^{2}. (2003  2 Marks)
Solution. (a) eV_{0} = hv – hv_{0} = 5 – 3 = 2 eV
∴ V_{0} = 2 volt
(b) Note : When the intensity is doubled, the saturation current is also doubled.
Q.32. A radioactive sample of ^{238}U decays to Pb through a process for which the halflife is 4.5×10^{9} years. Find the ratio of number of nuclei of Pb to ^{238}U after a time of 1.5×10^{9} years.
Given (2)^{1/3} = 1.26. (2004  2 Marks)
Ans. 0.26
Solution. a = Initial Uranium atom
(a – x) = Uranium atoms left
Q.33. The photons from the Balmer series in Hydrogen spectrum having wavelength between 450 nm to 700 nm are incident on a metal surface of work function 2 eV. Find the maximum kinetic energy of ejected electron. (Given hc = 1242 eV nm) (2004  4 Marks)
Ans. 0.55 eV
Solution. KEY CONCEPT :
The wavelength λ, of photon for different lines of Balmer series is given by
Using above relation, we get the value of λ = 657nm, 487 nm between 450 nm and 700 nm. Since 487 nm, is smaller than 657 nm, electron of max. K.E. will be emitted for photon corresponding to wavelength 487 nm with
Q.34. The potential energy of a particle of mass m is given by λ_{1} and λ_{2} are the deBroglie wavelengths of the particle, when 0 < x < 1 and x > 1 respectively. If the total energy of particle is 2E_{0}, find λ_{1} / λ_{2}. (2005  2 Marks)
Ans. √2
Solution. The de Broglie wave length is given by
Case (i) 0 < x < 1
For this, potential energy is E_{0} (given)
Total energy = 2E_{0} (given)
∴ Kinetic energy = 2E_{0} – E_{0} = E_{0}
Case (ii) x > 1
For this, potential energy = 0 (given)
Here also total energy = 2E_{0} (given)
∴ Kinetic energy = 2E_{0}
Dividing (i) and (ii)
Q.35. Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is (14)^{1/3}. Find (a) atomic number of the nucleus. (b) the frequency of K_{α} line of the Xray produced. (R = 1.1 × 10^{7} m^{–1} and c = 3 × 10^{8} m/s) (2005  4 Marks)
Ans. (a) 56
(b) 1.546 × 10^{18} Hz.
Solution. (a) KEY CONCEPT : We know that radius of nucleus is given by formula
r = r_{0}A^{1/3} where r_{0} = constt, and A = mass number.
For the nucleus r_{1} = r_{0}4^{1/3 }
For unknown nucleus r_{2} = r_{0} (A)^{1/3}
∴ No of proton = A – no. of neutrons = 56 – 30 = 26
∴ Atomic number = 26
Here, R = 1.1 × 10^{7}, c = 3 × 10^{8}, Z = 26
b = 1 (for K_{α}), n_{1} = 1, n_{2} = 2
Q.36. In hydrogenlike atom (z = 11), nth line of Lyman series has wavelength λ. The deBroglie's wavelength of electron in the level from which it originated is also λ. Find the value of n ? (2006  6M)
Ans. 24
Solution. Note : nth line of Lyman series means electron jumping from (n + 1)th orbit to 1st orbit.
For an electron to revolve in (n + 1)th orbit. 2πr = (n + 1)λ
Also we know that when electron jumps from (n + 1)th orbit to 1st orbit.
On solving, we get n = 24
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