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Published on: 02/11/2025
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1.
(a) Define the term decay constant and half life of a radioactive sample. Derive the relation connecting the two.
(b) How many disintegrations per second will occur in one gram of 92U238, if its half-life against alpha decay is \(1.42\times { 10 }^{ 17 }s\) ?
2.
What is the minimum energy that must b given to a H-atom in ground state so that it can angular momentum of the system is conserved, what would be the angular momentum of such H photon?
3.
What is Bohr's quantum condition?
4.
The wavelength of \({ K }_{ \alpha }\) line for copper is \(1.36\mathring { A } \) . Calculate the ionisation potential of a K shell electron in copper.
5.
Calculate shortest wavelength of Balmer series. Given \(R=1.097\times { 10 }^{ 7 }{ m }^{ -1 }\).
6.
Calculate the equivalent energy of electron and proton at rest.Given that mass of electron \(=9.1\times { 10 }^{ -31 }kg\) and mass of proton \(=1.673\times { 10 }^{ -27 }kg\) .
7.
The half-life of a certain radioactive material is 100 days. After how much time the undecayed fraction of a material will be 6.25%?
8.
One milligram of thorium emits 22 \(\alpha -particles\) per minute per unit solid angle. Calculate average life of thorium. The Atomic weight of thorium is 232.
9.
A radioactive sample contains 2.2 mg of pure \(_{ 6 }{ { C }^{ 11 } }\) which has half-life of 1224 s. Calculate
(i) the number of atoms presents initially
(ii) the activity when \(5\mu g\) of the sample will be left?
10.
Assuming that four hydrogen atoms combine to form a helium atom and two positrons, each of mass 0.000549 u, calculate the energy released. Given \(m(_{ 1 }{ { H }^{ 1 }) }\)= 1.007825 u and \(m(_{ 2 }{ { He }^{ 4 }) }\) = 4.002604 u.
11.
There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutrons is 700 seconds, what fraction of neutrons will decay before they travel a distance of 10 m? Given mass of neutron \(=1.675\times { 10 }^{ -27 }kg.\)
12.
Suppose you are given a chance to repeat the alpha particle scattering experiment using a thin sheet of solid hydrogen in place of gold foil. What results do you expect?
1.
\(1.23\times { 10 }^{ 4 }{ s }^{ -1 }\)
2.
Balmer series of \(H_\gamma\) required transition from n=5 to n = 1
So energy required to excite electron fromn=1 to n = 5 is
E = E_1E_5 = 13.6-0.54 = 13.06 eV
if angular momentum is conserved, then angular momentum of photon
= L5-L2 = 5\(\hbar\)-2\(\hbar\)= 3\(\hbar\)
= 3x1.06 x 10-34kg m2s-1
= 3.18 x 10-34 kg m2s-1
3.
According to Bohr's quantum condition, the permitted orbits are those in which the angular momentum of the electron is integral multiple of \(nh/2\pi\), where h is Planck's constant
i.e. m v r = \(nh/2\pi\) (when n = 1, 1, ............ called principal quantum number).
4.
\(1.22\times { 10 }^{ 4 }V\)
5.
\(3646.8\mathring { A } \)
\(Take \ { n }_{ 1 }=2 \ and \ { n }_{ 2 }=\infty \)
6.
0.511 MeV, 941.1 MeV
\({ E }_{ 1 }={ m }_{ e }{ c }^{ 2 }=9.1\times { 10 }^{ -31 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }J\)
\(=\frac { 81.9\times { 10 }^{ -15 } }{ 1.6\times { 10 }^{ -13 } } MeV=0.511 \ MeV\)
\({ E }_{ 2 }={ m }_{ p }{ c }^{ 2 }=1.673\times { 10 }^{ -27 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }J\)
\(=\frac { 1.673\times { 9\times 10 }^{ -11 } }{ 1.6\times { 10 }^{ -13 } } MeV=941.1 \ MeV\)
7.
400 days.
8.
\(1.782\times { 10 }^{ 10 }years\)
9.
\(1.2\times { 10 }^{ 20 };\ 1.55\times { 10 }^{ 14 }disintegrations/sec\)
10.
25.7 MeV
11.
\(Here; \ K.E.=\frac { 1 }{ 2 } { m\upsilon }^{ 2 }=0.0327 \ eV=0.0327\times 1.6\times { 10 }^{ -19 }J\)
\(\\ or \ \ \frac { 1 }{ 2 } \times \left( 1.675\times { 10 }^{ -27 } \right) \times { \upsilon }^{ 2 }=0.0327\times 1.6\times { 10 }^{ -19 }\)
\(On \ solving, \ we \ get \ \upsilon \ = \ 2.5\times { 10 }^{ 3 }m/s.\)
\(Time, \ t=\frac { distance }{ velocity } =\frac { 10 }{ 2.5\times { 10 }^{ 3 } } =4\times { 10 }^{ -3 }s\)
\(Now \ N={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/T }or \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ t/T }={ \left( \frac { 1 }{ 2 } \right) }^{ \frac { 4\times { 10 }^{ -3 } }{ 700 } }=0.999952\)
\(\therefore \ Fraction \ of \ neutron \ decayed \ = \ 1-0.999952 \ = \ 0.000048\)
12.
The basic purpose of scattering experiment is defeated because solid hydrogen will be much lighter target compared to the alpha particle acting as a projectile. According to the theory of elastic collisions, the target hydrogen will move much faster compared to alpha, after the collision. We cannot determine the size of hydrogen nucleus.
12th Standard CBSE Syllabus & Materials
12th Standard CBSE
CBSE 12th Chemistry d- and f- Block Elements Important Questions And Answers Study Material - QB365 Set A
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