Atomic Structure Question 475

Question: Photon having wavelength 310 nm is used to break the bond of $ A_2 $ molecule having bond energy $ 288,kg,mo{l^{-1}} $ then % of energy of photon converted to the K.E. is

$ [hc=12400,ev,\overset{o}{\mathop{A}},,,1,ev=96,kJ/mol] $

Options:

A) 25

B) 50

C) 75

D) 80

Show Answer

Answer:

Correct Answer: A

Solution:

  • Energy of on photon $ =\frac{12400}{3100}=4,eV= $

$ 4\times 96=384kJ,mo{l^{-1}} $

$ \therefore $ % of energy converted to $ K.E.=\frac{384-288}{384}=\frac{96}{384}\times 100=25% $



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