Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.
From E= hv or hc/ λ, we have λ = 242nm or 242 x 10-9m
C = 3 x 108 m/s
h = 6.62 x 10-34 Js
Now putting these values in the equation we get
E= (6.62 x 10-34 Js x 3 x 108 m/s) / (242 x 10-9m)
= 0.0821 x 10-17 J/atom
Or
E = (0.0821 x 10-17 ) / (1000 x 6.02 x 1023)
= 494 KJ/mol
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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 10: Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calcula....
Comments
Thq a lot to give this answers
I have problem in dealing with numerical
Thank
Y
What about if use hydrogen instead of sodium
Tq saralstudy
Why cant we use -13.6Z^2
Why cant we use -13.6Z^2
How it's come?
Right solution