SATHEE: Atomic Structure Question 426

Atomic Structure Question 426

Question: The dissociation energy of $H_{2}$ is $430.53 k J m o l^{- 1}$ . 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, $E = N_{0} \times h_{v}$

$= \frac{N_{0} \times h \times c}{λ}$

$= \frac{6.023 \times 10^{23} \times 6.63 \times 10^{- 34} \times 3 \times 10^{8}}{253.7 \times 10^{- 9}}$ Energy converted into $K E = (472.2 - 430.53) k J$ % of energy converted into $K E = \frac{(472.2 - 430.53)}{472.2} \times 100 = 8.82$ %

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Atomic Structure Question 426

Question: The dissociation energy of H2 is 430.53kJmol−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:

Answer:

Solution:

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Question: The dissociation energy of $H_{2}$ is $430.53 k J m o l^{- 1}$ . 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