SELECT * FROM question_mgmt as q WHERE id=1267 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=43 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
The electron energy in hydrogen atom is given by En = (–2.18 × 10–18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
The expression for the energy of hydrogen of electron is
En = -2π2me4Z2/n2h2
Where m= mass of electrons
Z=atomic mass of atom
e = charge of electron
h = planck’s constant
When n = 1 then En = - (2.18X10-18 ) (GIVEN)
When n = 2 then En= - (2.18X10-18 )/4 = 0.5465X10-18 J
Therefore the energy required for ionization from n = 2 is 5.45 x 10-19 J
Now wavelength of light needed
= E=hv = hc/λ
Or
λ = [{(6.62x10-24)(3x108)} / 5.45] x 10-19 = 3647
The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
Elements | ΔiH1 | ΔiH | ΔegH |
I | 520 | 7300 | -60 |
II | 419 | 3051 | -48 |
III | 1681 | 3374 | -328 |
IV | 1008 | 1846 | -295 |
V | 2372 | 5251 | +48 |
VI | 738 | 1451 | -40 |
Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX2, (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
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