Atomic Structure Question 295
Question: The dissociation energy of $ H_2 $ is $ 430.53kJmo{l^{-1}} $ . If hydrogen is dissociated by illumination with radiation of wavelength 253.7 nm the fraction of the radiant energy which will be converted into kinetic energy is given by
Options:
A) 100%
B) 8.82%
C) 2.22%
D) 1.22%
Show Answer
Answer:
Correct Answer: B
Solution:
- Energy of 1 mole of photons, $ E=N_0\times h_{v} $
$ =\frac{N_0\times h\times c}{\lambda } $
$ =\frac{6.023\times 10^{23}\times 6.63\times {10^{-34}}\times 3\times 10^{8}}{253.7\times {10^{-9}}} $ Energy converted into $ KE=( 472.2-430.53 )kJ% $ % of energy converted into $ KE=\frac{( 472.2-430.53 )}{472.2}\times 100=8.82% $ %
BETA
