Atomic Structure Question 355
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% $
BETA
