Test: Atomic Physics - 4 - JEE MCQ

 Therefore

Thus . When is doubled, becomes onefourth. Hence the correct choice is (d).

The longest wavelengths in the two series are given by

Now   Therefore, for hydrogen (n = 1 state), we have
r₁ = K     (∵Z = 1)

and for Li⁺⁺(n = 3 state) we have

Hence the correct choice is (c).

Given E₂ − E₁ = 10.2eV = 10.2 × 1.6 × 10⁻¹⁹ J Therefore, frequency the emitted radiation is

Energy of electron in hydrogen atom in ground state = −13−5eV.
When a photon of energy of 12.1eV is radiated on atom, the energy gained by the electron is given by:
E = (−13.6) + 12.1 = (−1.5)eV
And we know that energy of an excited electron is given by,

The distance of the closest approach is given by 
where . Thus When is doubled, becomes half. Hence the correct choice is (c).

The kinetic and potential energies of an electron in the th excited state are given by
KE  

Adding (i) and (ii) we get the total energy which is

Notice from (i) and (iii) that E = −KE. Given E = −3.4eV. Hence KE = −E = −(−3.4) = +3.4eV
Thus the correct choice is (b).

The total energy of an electron bound to an atom is negative and is the sum of its P.E. and K.E. The magnitude of the P.E. is twice that of the K.E. (as per the Bohr model) but since the P.E. is -ve, we have Total Energy = (−2K) + (K) = −K (K = Kinetic energy). Hence the ratio of total energy and kinetic energy = −1:1 = −1

The number of scintillations per unit time is  Hence


Hence  which is choice (a).

From (ii) and (iii) we find that .
Hence the correct choice is (c).